■ 


UC-NRLF 


*B    S2fl    337 


h 

LIBRARY 

OF  THE 

University  of  California. 

GIFT  OF 

C7#ss 

TO  EDUCATOE8. 


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WILSON,  HINKLE  &  CO., 

CINCINNATI   AND  NEW   YORK. 


Ray's   Series, 


EMBRACING 


A  Thorough  and  Progressive  Course  in  Arithmetic,  Algebra, 
and  the  Higher  Mathematics. 


Primary  Arithmetic. 
Intellectual  Arithmetic. 
Rudiments  of  Arithmetic. 


Higher  Arithmetic. 

Test  Examples  in  Arithmetic. 

New  Elementary  Algebra. 


Practical  Arithmetic.  New  Higher  Algebra. 


Plane  and  Solid  Geometry.    By  Eli  T.  Tappan,  A.M.,  Prcs't 

Kenyon  College.     12mo,  cloth,  276  pp. 
Geometry    and    Trigonometry.     By   Eli   T.   Tappak,  A.M. 

Prcs't  Kenyon    College.     8vo,  sheep,  420  pp. 
Analytic  Geometry,     By  Geo.  H.  Howisok,  A.M.,  Prof,  in  Mass. 
Institute  of  Technology.    Treatise  on  Analytic  Geometry,  especially 
as  applied  to  the  Properties  of   Conies :    including  the  Modern 
Methods  of  Abridged  Notation. 

Elements  of  Astronomy.    By  S.  H.  Peabody,  A.M.,  Prof,  of 

Physics  and  Civil  Engineering,  Amherst  College.      Handsomely  and 
profusely  illustrated.     &vo,  sheep,  336  pp. 


KEYS. 

Ray's  Arithmetical  Key  ( To  Intellectual  and  Practical)  ; 

Key  to  Kay's  Higher  Arithmetic ; 

Key  to  Ray's  Sew  Elementary  and  Higher  Algebras. 


Descriptive   Circulars   and   Price  T^ist    upon  Apj>lication. 


WILSON,  HINKLE  &  CO.,  PUBLISHERS, 

CINCINNATI:  NEW    YORK 

137  Walnut  St.  28  Bond  St. 


EVANS'    SCHOOL     GEOMETRY. 


PRIMARY  ELEMENTS 


OF 


PLANE  AND  SOLID   GEOMETRY: 


FOR  SCHOOLS  AND  ACADEMIES. 


By  E.  W.  EVANS,  M.  A., 

PBOrtSSOR     OF      MATHEMATICS      IN     MARIETTA     COLIEOE. 


WILSON,   HINKLE   &  CO., 


CIISrCINlNr^TI: 

137  "Walnut  St. 


NEW    YORK 
28  Bond  St. 


Entered  according  to  Act  of  Congress,  in  the  year  1862,  by 

"W.  B.  SMITH  &  CO., 

In  the  Clerk's  Office  of  the  District  Court  of  the  United  States,  for  the 
Southern  District  of  Ohio. 


Electrotyped  at  the 

Franklin  Type  Foundry, 

Cincinnati,  0. 


PREFACE. 


The  following  concise  treatise  on  Geometry  has  been 
prepared  chiefly  for  that  class  of  students  in  our  Pub- 
lic Schools  and  elsewhere,  who  can  not  spare  the  requisite 
time  for  mastering  the  larger  works.  It  is  conceived, 
also,  that  it  will  be  found  useful  as  a  course  of  first 
lessons  for  those  who  intend  to  pursue  the  study  more 
at  length. 

The  writer  has  aimed,  in  general,  to  select  those  prop- 
ositions in  both  Plane  and  Solid  Geometry  which  have 
the  most  important  direct  applications  ;  omitting  a  large 
number  of  those  which  serve  chiefly  as  steps  to  some- 
thing following  them  in  an  extended  course  of  reasoning. 
In  order  to  do  this  without  breaking  the  logical  con- 
nection, he  has  often  found  it  necessary  to  present  the 
demonstrations  in  a  considerably  modified  form  ;  though, 
in  a  majority  of  cases,  they  are  essentially  based  on 
those  found  in  Euclid,  Legendre,  and  other  standard 
works. 

In  the  treatment  of  parallel  lines,  important  hints 
have  been  taken  from  the  work  of  Professor  Peirce, 


6  PKEFACE. 

whose  method  on  this  subject  is  the  most  simple  and 
logical.  Upon  the  whole,  the  demonstrations  have  been 
put  in  that  form  which  was  thought  best  suited  to  the 
comprehension  of  beginners  having  no  previous  knowl- 
edge of  mathematics  beyond  arithmetic. 

In  the  divisions  of  the  work,  the  writer  has  aimed 
to  follow  the  simple  and  natural  classification  of  the 
subjects  treated.  The  arrangement  adopted  has  enabled 
him  to  set  down  as  simple  corollaries  many  propositions 
which  are  usually  demonstrated  separately.  The  defini- 
tions have  been  distributed  between  the  various  sections, 
instead  of  being  crowded  together  at  the  beginning. 
After  each  Section,  or  Book,  there  will  be  found  a  few 
practical  illustrations  and  exercises.  In  a  Supplement, 
some  examples  have  been  given  of  the  application  of 
Algebra  to  Geometry. 

The  amount  of  Geometry  contained  in  the  work  is 
sufficient  to  prepare  the  pupil  for  the  study  of  Plane 
Trigonometry  and  Surveying. 


CONTENTS. 


INTRODUCTION. 


PACK 


Sec.  I.  General  Definitions 9 

Sec.  IT.        Of  Terms  and  Signs 10 

Sea  III       Axioms H 

BOOK   I 
PLANE    GEOMETRY.  — RATIOS. 


Sec.  IV.       Straight  Lines  and  their  Angles 13 

Sec.  V.         Triangles 13 

Sec.  VI.       Quadrilaterals 2G 

Sec.  VII.     Of  Polygons  in  General 35 

See.  VIII.  Of  the  Circle 38 

Sec.  IX.       Problems  in  Construction 49 

* 

BOOK   II. 
PLANE     GEOMETRY.—  PROPORTIONS. 

Sec.  X.         Of  Polygons 55 

Sec.  XI.       Of  Circles 03 

Sec.  XII.     Problems  in  Construction 73 

(7) 


8  CONTENTS 

BOOK   III. 
SOLID    GEOMETRY. 

PAGE 

Sec.   XIII.     Planes  and  their  Inclinations 76 

Sec.  XIV.      Of  Parallelopipeds 79 

Sec.  XV.       The  Prism  and  the  Cylinder 83 

Sec.  XVI.     Pyramids  and  Cones 86 

Sec.  XVII.    The  Sphere 92 


SUPPLEMENT. 

Sec.  XVIII.   Miscellaneous  Examples 97 

Sec.  XIX.      Applications  of  Algebra .'...     99 


INTRODUCTION. 


SECTION    I—  GENERAL    DEFINITIONS. 

1.  Geometry  is  that  branch  of  mathematics  which 
treats  of  magnitudes  in  space. 

2.  A  line  is  that  magnitude  which  has  length,  with- 
out breadth  or  thickness. 

3.  A  surface  is  that  which  has  length  and  breadth, 
without  thickness. 

4.  A  solid  is  that  which  has  length,  breadth,  and 
thickness. 

5.  A  point  has  position  without  extension. 

6.  A  straight  line  is  one  which  has  the  same  di- 
rection through  all  its  consecutive  parts. 

7.  Of  bent  lines,  that  which  is  composed  of  straight 
lines  is  called  a  broken  lirie;  that  of  which  no  part  is 
straight  is  called  a  curve. 

8.  A  plane  is  a  surface  in  which  if  any  two  points 
whatever  be  taken,  the  straight  line  joining  them  will 
lie  wholly  in  that  surface. 

9.  Magnitudes  of  extension  are  lines,  surfaces,  and 
solids.  The  relative  positions  of  these  give  rise  to 
magnitudes  of  direction,  or  angles,  which  will  be  defined 
in  their  proper  connections. 

(9) 


10  GEOMETRY. 

SEC.  II.— OF    TEEMS    AND   SIGNS. 

1.  An  anjom  is  a  self-evident  truth. 

2.  A  theorem  is  a  statement  of  a  truth  which  re- 
quires to  be  proved,  or  demonstrated. 

8.  A  problem  is  a  statement  of  something  required 
to  be  done,  or  solved. 

4.  The  term  proposition  may  be  applied  either  to 
a  theorem  or  to  a  problem. 

5.  A  corollary  is  an  obvious  consequence  from 
something  that  precedes. 

6.  A  scholium  is  some  remark  relating  to  what 
precedes. 

7.  The  term  hypothesis  denotes  the  supposition 
made,  or  the  conditions  given,  in  any  proposition. 

8.  The  term  infinite,  as  used  in  geometry,  means 
beyond  measure ;  that  is,  either  absolutely  beyond 
limits,  or  beyond  all  appreciable  limits. 

9.  A  ratio  is  the  relation  which  one  quantity  bears 
to  another,  as  equal  to  it,  greater,  or  less.  The  value 
of  the  ratio  is  the  quotient  arising  from  dividing  one 
of  the  quantities  by  the  other.  If  A  and  B  represent 
any  two  quantities,  the  ratio  of  A  to  B  may  be  written 

A 

either  A  :  B,  or  — • 

Jj 

10.  A  proportion  is  an  equality  of  ratios.  Thus, 
if  the  ratio  of  A  to  B  be  equal  to  the  ratio  of  C  to 
D,  those  two  ratios  will  constitute  a  proportion,  which 
may  be  written  A  :  B  :  :  C  :  D. 

11.  The  sign  =  denotes  the  equality  of  two  quan- 
tities between  which  it  is  placed.     Hence, 

A        C 

A  proportion  may  also  be  written  —  =  — • 


INTRODUCTION.  11 

12.  The  sign  -f"  denotes  addition.  Thus,  A-\-B 
means  the  sum  of  A  and  B,  and  is  read  A  plus  B. 

13.  The  sign  —  denotes  subtraction.  Thus,  A — B 
means  A  diminished  by  B,  and  is  read  A  minus  B. 

14.  The  sign  X  denotes  multiplication.  Thus,  AxB 
means  A  multiplied  by  B.  Sometimes,  however,  this 
sign  is  omitted,  especially  if  one  of  the  factors  be  a 
figure.     Thus,  2B  means  twice  B. 

15.  The    sign    -r-    denotes    division.     Thus,    A-=-B 

means  A  divided  by  B.     It  is  equivalent  to  — 

16.  The  parenthesis  (  )  denotes  that  the  quantities 
inclosed  within  it  are  to  be  subjected  to  the  same 
operation.  Thus,  Ax(B-f-C)  means  A  multiplied  into 
the  sum  of  B  and  C. 

17.  The  square  of  any  quantity,  as  A,  is  written 
A2;  its  cube  is  written  A3. 

18.  The  square  root  of  any  quantity,  as  A-|-B,  is 
written  ^/A+B. 

SEC.     III.— AXIOMS. 

There  are  certain  axioms,  relating  to  quantity  in 
general,  which  lie  at  the  foundation  of  all  branches 
of  mathematics.  Others  relate  particularly  to  magni- 
tudes in  space.  Some  of  those  axioms,  of  both  classes, 
which  are  most  frequently  applied  in  geometry,  are 
stated  here. 

1.  The  whole  is  greater  than  any  of  its  parts. 

2.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

3.  Things  which  are  equal  to  the  same  thing  are 
equal  to  each  other. 

4.  If  equals  be  added  to  equals,  the  sums  will  be 
equal. 


12  GKOMETBY. 

5.  If  equals  be  taken  from  equals,  the  remainders 
will  be  equal. 

6.  If  equals  be  added  to  unequals,  the  sums  will  be 
unequal.  + 

7.  If  equals  be  taken  from  unequals,  the  remainders 
will  be  unequal. 

8.  Doubles  of  the  same   thing  are  equal   to    each 
other. 

9.  Halves  of  equal  things  are  equal  to  each  other. 

10.  From  one  point  to  another  only  one  straight 
line  can  be  drawn ;  and  that  is  the  shortest  line  be- 
tween them. 

11.  Two  magnitudes  are  equal  if,  when  one  is  ap- 
plied to  the  other,  they  will  exactly  coincide. 


r. 


Of  THE     Y 


UN/VERsrp. 


OF 

RNM 


PLANE   GEOMETRY. 


BOOK    I, 


RATIOS  OF  MAGNITUDES  LYING  IN  THE  SAME  PLANE. 


SEC.  IV.— STRAIGHT  LINES  AND  THEIR  ANGLES. 

DEFINITIONS. 

1.  The  divergence  of  two  straight  lines  from  a  point 
constitutes  an  angle.  The  quantity  of  the  angle  is 
the  difference  of  direction  between  the  two  lines. 

The  lines  themselves  are  said  to  contain  the  angle, 
and  are  called  its  sides.  The  point  from  which  they 
diverge  is  called  the  vertex.  .  C 

Thus,  AB  and  BC  are  the  sides,  and 
B  the  vertex  of  the  angle  ABC.  B^- A 

In  reading  angles  by  three  letters,  the  one  at  the 
vertex  should  always  occupy  the  middle  place. 

2.  When  the  adjacent  angles 
made  by  one  straight  line  with 
another  are  equal,  they  are  called 

right    angles.      Such     are    the     A 

angles  ACD  and  BCD. 

An  angle  which  is  greater  than 
a  right  angle  is  termed  OBTUSE  ;  an  angle  less  than  a 
right  angle,  acute. 

3.  One  straight  line  is  said  to  be  perpendicular  to 
another  when  it  makes  right  angles  with  it ;  oblique, 
when  it  makes  unequal  angles. 

(13) 


-B 


14  GEOMETRY. 

4.  Two  straight  lines  are  said  to  be  parallel  to 
each    other   when   they    have    the 

same    direction.      Thus,    AB    and 
CD  are  parallel. 

Corollary.  Parallel  lines  will  never  meet,  hoAV- 
ever  far  produced;  for,  if  they  were  to  meet,  they 
would  make  an  angle  with  each  other,  that  is,  they 
would  differ  in  direction,  which  is  contrary  to  the 
definition. 

Scholium.  Lines  having  the  same  direction  may 
also  be  said  to  have  exactly  opposite  directions  when 
taken  in  different  ways.  Thus,  the  direction  AB  is 
opposite  to  the  direction  DC. 

5.  A  straight  line,  or  any  other  magnitude,  is  said 
to  be  bisected  when  it  is  divided  into  two  equal 
parts. 

THEOREM    I. 

The  two  angles  which  one  straight  line  makes  with 
another,  on  one  side  of  it,  are  together  equal  to  two  right 
angles. 

Let  the  straight  line  DC 
meet  the  straight  line  AB  in 
C.  Then  will  the  two  angles 
ACD,  DCB,  be  together  equal 
to  two  right  angles.  ° 

For,  if  DC  be  perpendicular  to  AB,  each  of  these 
angles  will  be  a  right  angle  (Definition  8,  Section  IV). 
But  if  DC  is  oblique  to  AB,  erect  the  perpendicular 
CE.  Now,  the  part  ACE  of  the  obtuse  angle  ACD, 
is  a  right  angle ;  and  the  remaining  part  ECD  and  the 


BOOK    I. 


15 


acute  angle  DCB  together  make  up  another  right 
angle,  ECB. 

Therefore,  the  two  angles  which  one  straight  line 
makes  with  another,  on  one  side,  are  together  equal  to 
two  right  angles. 

Cor.  1.  In  a  similar  manner,  whatever  be  the  num- 
ber of  adjacent  angles  formed  at  C,  on  one  side  of  AB, 
it  may  be  shown  that  their  sum  is  two  right  angles. 

Cor.  2.  A  right  angle  is  the  fourth  part  of  the 
ivhole  compass  in  the  plane  about  any  point,  as  C ; 
for,  half  of  this  compass  lies  on  one  side  of  AB,  and 
half  on  the  other. 


THEOREM    II. 

If  two  straight  lines  cut  one  another,  the  opposite  or 
vertical  angles  are  equal. 

Let  AB  and  CD  be  two 
straight  lines  cutting  one  an- 
other in  E.  Then  will  AEC 
be  equal  to  its  opposite  or  ver- 
tical angle  DEB. 

For  the  angles  AEC,  AED,  on  one  side  of  CD,  are 
together  equal  to  two  right  angles  (Theo.  I) ;  so,  also, 
the  angles  AED,  DEB,  on  one  side  of  AB,  are  together 
equal  to  two  right  angles ;  therefore,  the  sum  of  AEC 
and  AED  is  equal  to  the  sum  of  AED  and  DEB 
(Axiom  3).  Now  take  away  the  common  angle  AED, 
and  the  remaining  angle  AEC  is  equal  to  the  remain- 
ing angle  DEB  (Ax.  5). 

In  the  same  manner  it  may  be  proved  that  the 
angle  AED  is  equal  to  its  opposite  angle  CEB. 

Therefore,  if  two  straight  lines,  etc. 


16 


GEOMETRY. 


•vl^\ 


V 


THEOREM    III. 

If  a  straight  line  intersect  two  parallels,  the  corres- 
ponding inner  and  outer  angles  will  be  equal  to  each 
other;  also  the  alternate  angles. 

Let  the  straight  line  EF 
intersect  the  two  parallel 
straight  lines  AB,  CD,  in 
G  and  H.  Then  will  any 
two  corresponding  inner  and 
outer  angles,  as  GHD  and 
EGB,  be  equal.  ■— 

For,  since  EF  is  a  straight  line,  its  part  HG  has 
the  same  direction  from  H  that  its  part  GE  has  from 
G  (Def.  G,  Sec.  I)  ;  and  since  HD  and  GB  are  by  hy- 
pothesis parallel,  HD  has  the  same  direction  from  H 
that  GB  has  from  G  (Def.  4,  Sec.  IV).  Therefore,  the 
difference  in  direction  between  HG  and  HD  is  equal 
to  the  difference  in  direction  between  GE  and  GB ; 
that  is,  the  angle  GHD  is  equal  to  the  angle  EGB 
(Def.  1,  Sec.  IV). 

A^ain,  any  two  alternate  angles,  as  AGH  and 
GHD,  are  equal  to  each  other.  For  AGH  is  equal 
to  its  opposite  angle  EGB  (Theo.  II).  But  EGB  has 
just  been  proved  equal  to  GHD.  Consequently,  AGH 
is  equal  to  GHD  (Ax.  3). 

Therefore,  if  a  straight  line  intersect,  etc. 

Cor.  1.  It  is  evident  that  if  AB  is  not  parallel  to 
CD,  but  takes  some  other  direction,  as  mGn,  the  cor- 
responding inner  and  outer  angles  will  be  unequal; 
also,  the  alternate  angles. 

Cor.  2.     If  two    angles    have    their    sides    parallel, 


BOOK    I. 


17 


each  to  each,  and  directed  the  same  way  from  the  ver- 
tex, they  are  equal. 


.F* 


THEOREM   IV. 

If  a  straight  line  intersect  two  parallels,  the  sum  of 
the  two  inner  angles  on  the  same  side  ivill  be  equal  to 
two  right  angles. 

Let  EF  intersect  the  two 
parallels  AB,  CD,  in  G  and 
H.  Then  will  BGH  and 
GHD  be  together  equal  to 
two  right  angles. 

For  the  sum  of  the  an- 
gles GHC  and  GHD  is  two 
right  angles  (Theo.  I).  But  GHC  is  equal  to  its 
alternate  angle  BGH  (Theo.  III).  Therefore,  the  sum 
of   BGH    and    GHD   is    equal    to    two    right    angles. 

Hence,  if  a  straight  line,  etc. 

Cor.  1.  If  BGH  is  a  right  angle,  GHD  must  also 
be  a  right  angle.  Hence,  if  a  straight  line  is  perpen- 
dicular to  one  of  two  parallels,  it  is  perpendicular  to 
the  other. 

Cor.  2.  If  two  straight  lines  are  both  perpendicular 
to  a  third,  they  are  parallel. 


EXERCISES. 

1.  Prove  that  the  sum  of  all  the  adjacent  angles 
made  by  any  number  of  straight  lines  meeting  in  one 
point  is  equal  to  four  right  angles. 

2.  Prove   that   if  one    of  the    four  angles  made  by 

Evans?  Geometry. — "2 


18 


GEOMETR  Y. 


two    straight    linos  intersecting  each  other  be  a  right 

O  O  O 

angle,  each  of  the  others  will  be  a  right  angle. 

8.  Prove  that  the  alternate  outer  angles  EGA  and 
FIID  (Figure  to  Theo.  Ill)  are  equal  to  each  other. 

4.  Prove  that  the  sum  of  the  two  outer  angles  on 
the  same  side,  EGB  and  FHD  (Fig.  Theo.  IV),  is  equal 
to  two  right  angles. 

5.  Point  out  all  the  angles  equal  to  EGB  (Fig. 
Theo.  IV) ;  also,  all  the  angles  equal  to  EGA ;  and 
show  in  each  case  why  they  are  equal. 


SEC.  V.— TRIANGLES. 

DEFINITIONS. 

1.  A  plane  figure  is  a  portion  of  a  plane  bounded 
on  all  sides  by  lines. 

2.  A  polygon  is  a  plane  figure  bounded  by  straight 
lines. 

3.  A  triangle  is  a  polygon  of  three  sides. 

If  it  has  one  right  angle  it  is 
called  a  right-angled  triangle.  The 
side  opposite  the  right  angle  is 
called  the  hypotenuse. 


A  triangle  which  has  no  right 
angle  is  called  an  oblique-angled 
triangle. 


An  isosceles  triangle  is  one  which 


has  two  equal  sides. 


BOOK    I.  19 

4.  Any  side  of  a  triangle  may  be  considered  as  its 
base,  and  the  opposite  angle  as  its  vertex;  but  in  an 
isosceles  triangle,  that  side  is  usually  called  the  base 
which  is  not  equal  to  either  of  the  others. 

The  altitude  of  a  triangle  is  the  perpendicular  let 
fall  from  the  vertex  on  the  base,  or  the  base  produced. 

5.  A  triangle,  or  other  polygon,  is  called  equiangu- 
lar when  all  its  angles  are  equal ;  equilateral,  when  all 
its  sides  are  equal. 

6.  Two  polygons  are  called  mutually  equilateral  or 
mutually  equiangular,  if  the  sides  or  angles  of  the  one 
are  equal  to  the  sides  or  angles  of  the  other,  each  to 
each,  taken  in  the  same  order. 

THEOREM    V. 

Tlie  three  interior  angles  of  any  triangle  are  together 
equal  to  two  right  angles. 

Let  ABC  be  any  tri- 
angle. It  is  to  be  proved 
that  the  angles  ABC,  BCA, 

CAB,    are    together  equal  ^/_  _D 

to  two  right  angles. 

Produce  AB  to  any  point  D,  and  draw  BE  parallel 
to  AC.  Now,  the  sum  of  the  three  angles  ABC,  CBE, 
EBD,  is  equal  to  two  right  angles  (Cor.  1,  Theo.  I). 
But  because  CB  intersects  the  parallels  AC  and  BE, 
the  angle  CBE  is  equal  to  its  alternate  angle  BCA 
(Theo.  Ill)*;  also,  because  AD  intersects  the  same  par- 
allels, the  angle  EBD  is  equal  to  its  corresponding 
inner  angle  CAB.  Therefore,  the  sum  of  ABC,  BCA, 
CAB,  is  equal  to  two  right  angles. 

That  is,  the  three  interior  angles  of  a  triangle,  etc. 


20 


GEOMETRY. 


Cor.  1.  If  one  side  of  a  triangle  be  produced,  the 
exterior  angle  ivill  be  equal  to  the  sum  of  the  two  oppo- 
site interior  angles.  For  the  sura  of  CBE  and  EBD, 
that  is,  the  whole  exterior  angle  CBD,  is  equal  to  the 
sum  of  BCA  and  CAB. 

Cor.  2.  A  triangle  can  not  have  more  than  one  rigid 
angle;  for,  if  it  had  two,  the  third  angle  would  be 
nothing.  Still  less  can  a  triangle  have  more  than  one 
obtuse  angle. 


THEOREM   VI. 

If  two  straight  lines  be  drawn  from  the  extremities 
of  one  side  of  a  triangle  to  a  point  within,  their  sum 
will  be  less  than  that  of  the  other  two  sides  of  the 
triangle. 

From  the  extremities  of  AB,  let 
straight  lines  be  drawn  to  a  point 
D  within  the  triangle  ABC.  It  is 
to  be  proved  that  the  sum  of  AD 
and  DB  is  less  than  the  sum  of 
AC  and  CB. 

Produce  AD  to  meet  BC  in  E.  Now,  since  AE  is 
a  straight  line,  it  is  less  than  the  sum  of  AC  and 
CE,  which  form  a  broken  line  (Ax.  10).  Therefore, 
if  we  add  EB  to  both,  it  is  evident  that  the  sum.  of 
AE  and  EB  is  less  than  the  sum  of  AC,  CE,  and 
EB,  that. is,  less  than  the  sum  of  AC  and  CB  (Ax.  6). 
In  the  same  manner  it  may  be  shown  that  the  sum 
of  AD  and  DB  is  less  than  the  sum  of  AE  and  EB ; 
still  more,  then,  we  may  conclude  it  is  less  than  the  sum 
of  AC  and  CB. 

Therefore,  if  two  straight  lines,  etc. 


BOOK    I.  21 

THEOREM    VII. 

If  two  triangles  have  two  sides  and  the  included 
angle  of  the  one  equal  to  two  sides  and  the  included 
angle  of  the  other,  each  to  each,  they  are  equal  through- 
out. 

Let  the  triangles  ABC,  DEF,  have  the  side  AB 
equal  to  the 
side  DE,  and 
AC  equal  to 
DF,  and  the 
included  angle 
A    to   the    in- 


B  D  "  E 


eluded  angle  D.  Then  will  the  two  triangles  be  equal 
throughout;  that  is,  the  third  side  BC  will  be  equal 
to  the  third  side  EF,  the  angle  B  to  the  angle  E,  the 
angle  C  to  the  angle  F,  and  the  triangle  ABC  as  a 
whole  to  the  triangle  DEF  as  a  whole. 

If  the  triangle  ABC  be  applied  to  the  triangle  DEF 
so  that  the  point  B  shall  be  upon  the  point  E,  and  the 
side  BA  upon  the  side  ED,  then  since  these  sides  are 
by  hypothesis  equal,  the  point  A  will  fall  upon  the 
point  D  ;  and  since  the  angle  A  is  equal  to  the  angle 

D,  the  side  AC  will  fall  upon  the  side  DF  ;  and  since 
these  sides  again  are  equal,  the  point  C  will  fall  upon 
the  point  F.     But  the  point  B  is  also  upon  the  point 

E.  Therefore,  the  side  CB  must  coincide  with  the 
side  FE  (Ax.  10),  and  the  angle  C  with  the  angle  F, 
and  the  angle  B  with  the  angle  E,  and  the  triangle 
ABC  as  a  whole  with  the  triangle  DEF  as  a  whole ; 
and  since  they  coincide,  each  with  each,  they  are  also 
equal  (Ax.  11).        ,  -    '      ■ 

Therefore,  if  two  triangles,  etc. 


22 


GEOMETRY. 


Cor.     In  equal  triangles,  equal  angles  are  opposite 
to  equal  sides. 


THEOREM    VIII. 

The  angles  at  the  base  of  an  isosceles  triangle  are 
equal  to  each  other. 

Let  ABC  be  an  isosceles  triangle 
having  the  side  AC  equal  to  the  side 
BC.  It  is  to  be  proved  that  the  angle 
A  is  equal  to  the  angle  B. 

Draw  CD  bisecting  the  angle  ACB. 
Then  in  the  triangles  ADC,  BDC,  A"  "^ 
since  the  side  AC  is,  by  hypothesis,  equal  to  the  side 
BC,  and  the  side  CD  is  common  to  both,  and  the  in- 
cluded angle  ACD  is,  by  construction,  equal  to  the 
included  an^le  BCD,  it  follows  that  the  triangle  ADC 
is  equal  to  the  triangle  BDC  (Theo.  VII),  and  the 
angle  A  to  the  angle  B  (Cor.  Theo.  VII). 

Therefore,  the  angles  at  the  base,  etc. 

Cor.  1.  A  straight  line  bisecting  the  vertical  angle 
of  an  isosceles  triangle  is  a  perpendicular  to  the  middle 
point  of  the  base.  For  AD  is  equal  to  DB  ;  also,  the 
angles  CDA,  CDB,  being  equal,  are  right  angles  (Def. 

2,  Sec.  4). 

Cor.  2.  An  equilateral  triangle  is  also  equian- 
gular. 


THEOREM    IX. 

The  greater   of  two   unequal  sides  in  a  triangle  has 
the  greater  angle  opposite  to  it. 


BOOK    I 


23 


Let  ABC  be  a  triangle  having  the 
side  AB  greater  than  the  side  BC. 
Then  will  the  angle  BGA  be  greater 
than  the  angle  BAC. 

Produce  BC  so  as  to  make  BD 
equal  to  BA ;  also,  join  DA.  Now, 
in  the  isosceles  triangle  ABD,  the  A 
angles  BDA,  BAD,  are  equal  (Theo.  VIII).  But  the 
exterior  angle  BCA,  of  the  triangle  ACD,  is  greater 
than  the  opposite  interior  angle  BDA  (Cor.  I,  Theo. 
V),  and  consequently  greater  than  BAD ;  still  more, 
then,  is  it  greater  than  BAC. 

Therefore,  the  greater  of  two  unequal  sides,  etc. 

Cor.     Hence,  if  two  angles  of  a  triangle  are  equal, 
the  sides  opposite  to  them  can  not  be  unequal. 


THEOREM   X. 

If  two  triangles  have  hvo  sides  of  the  one  equal  to 
two  sides  of  the  other,  but  the  included  angles  unequal, 
the  base  of  that  which  has  the  greater  angle  is  greater 
than  the  base  of  the  other. 

Let  the  two  tri- 


angles ABC,  DEF, 
have  the  two  sides 
AB,  AC,  respect- 
ively equal  to  the 
two  sides  DE,  DF,  but  the  angle  BAC  greater  than 
the  angle  EDF.  It  is  to  be  proved  that  the  base  BC 
is  greater  than  the  base  EF. 

Because  the  angle  EDF  is,  by  hypothesis,  less  than 
the  angle  BAC,  the  sum  of  the  angles  E  and  F  must 
be  greater  than  the  sum  of  the  angles  B  and  C  (Theo. 


24  GEOMETRY. 

V) :  hence,  either  E  is  greater  than  B,  or  F  is  greater 
than  C.  Suppose,  then,  that  F  is  greater  than  C. 
On  the  adjacent  side  DF,  describe  the  triangle  DGF 
equal  to  the  triangle  ABC,  and  so  placed  that  DG 
shall  be  the  side  equal  to  AB.  Now,  the  sum  of  DH 
and  HG  is  greater  than  DGL  (Ax.  10) ;  and  the  sum 
of  EH  and  HF  is  greater  than  EF.  Therefore,  the 
sum  of  the  entire  lines  DE  and  GF  is  greater  than 
the  sum  of  DG  and  EF,  Taking  away  from  these 
unequals  the  equals  DE  and  DG,  we  have  the  remain- 
der GF  greater  than  the  remainder  EF.  But  GF  is 
equal  to  BC  by  construction.  Therefore,  BC  is 
greater  than  EF. 

Hence,  if  two  triangles,  etc. 

THEOREM  XL 

If  two  triangles  have  two  angles  and  the  included 
side  of  the  one  equal  to  two  angles  and  the  included 
side  of  the  other,  each  to  each,  they  are  equal  throughout. 


C  F 


/ 


Let  ABC,  DEF, 
be  two  triangles 
having  the  angles 
A  and  B  respect-     .X  _\B      -^/_  \.£ 

ively  equal  to  the 

angles  D  and  E,  and  the  included  side  AB  equal  to 
the  included  side  DE.  It  is  to  be  proved  that  these 
triangles  are  equal  throughout, 

If  the  triangle  ABC  be  applied  to  the  triangle  DEF, 
so  that  the  side  AB  shall  coincide  with  its  equal  DE, 
then  because  the  angle  A  is  equal  to  the  angle  D,  the 
side  AC  will  fall  on  the  side  DF,  and  the  point  C  will 
be  found  somewhere  in  the  line  DF ;  also,  because  the 


BOOK    I. 


25 


angle  B  is  equal  to  the  angle  E,  the  side  BC  will  fall 
on  the  side  EF,  and  the  point  C  will  be  found  some- 
where in  the  line  EF ;  consequently  the  point  C,  being 
in  both  the  lines  DF  and  EF,  must  be  at  their  point 
of  intersection  F.  Hence,  the  two  triangles  exactly 
coincide,  and  are  equal  throughout  (Ax.  11) ;  that  is, 
the  side  AC  is  equal  to  the  side  DF,  the  side  BC  to 
the  side  EF,  the  angle  C  to  the  angle  F,  and  the  tri- 
angle ABC  as  a  whole  to  the  triangle  DEF  as  a  whole. 

Therefore,  if  two  triangles,  etc. 


THEOKEM   XII. 

Two  triangles  which  are  mutually  equilateral  are  also 
mutually  equiangular. 

Let  the  two  tri-  A 

angles  ABC,  DEF, 
have  the  three  sides 
AB,  BC,  CA,  re- 
spectively equal  to 
the  three  sides  DE,  EF,  FD.  Then,  also,  will  the 
angles  A,  B,  and  C  be  respectively  equal  to  the  angles 
D,  E,  and  F. 

If  the  angle  A  is  not  equal  to  the  angle  D,  it  must 
be  either  greater  or  less.  But  it  can  not  be  greater 
than  D,  for  then  the  base  BC  would  be  greater  than 
the  base  EF  (Theo.  X),  which  is  contrary  to  the  hy- 
pothesis. Neither  can  it  be  less  than  D  ;  for,  in  that 
case,  BC  would  be  less  than  EF,  which  is  also  con- 
trary to  the  hypothesis.  Hence,  A  must  be  equal  to 
D  :  and,  in  the  same  manner,  it  may  be  proved  that  B 
is  equal  to  E,  and  C  to  F. 

Therefore,  two  triangles,  etc. 

Evans    Geometry. — 3  ^ 


26  GEOMETRY. 

Cor.  If  two  triangles  are  mutually  equilateral  they 
are  equal  throughout. 

EXERCISES. 

1.  Prove  that  if  two  angles  of  one  triangle  are 
equal  to  two  angles  of  another,  the  third  angles  arc 
equal. 

2.  Prove  that  each  angle  of  an  equilateral  triangle 
is  equal  to  two-thirds  of  a  right  angle. 

3.  Prove  that  if  a  perpendicular  be  erected  on  the 
middle  point  of  a  straight  line,  any  point  in  it  will 
be  equally  distant  from  the  extremities  of  that  line 
(Theo.  VII). 

4.  Prove  that  from  a  given  point  without  a  straight 
line,  only  one  perpendicular  to  that  line  can  be  drawn 
(Cor.  2,  Theo.  V). 

5.  If  the  vertical  angle  of  an  isosceles  triangle  be 
two-sevenths  of  a  right  angle,  what  will  be  the  value 
of  each  of  the  angles  at  the  base  ? 


SEC.  VI.— QUADRILATERALS. 
DEFINITIONS. 

1.  A  quadrilateral  is  a  polygon  of  four  sides. 

2.  A  parallelogram  is  a  quad- 


rilateral   having  its  opposite  sides 

parallel.  L j 

A  parallelogram  whose  angles 
are  all  right  angles,  is  called  a 
rectangular  parallelogram,  or  sim- 
ply a  RECTANGLE, 


BOOK    I.  27 


A  SQUARE  is  a  rectangle  whose 
sides  are  all  equal. 

3.  A  trapezoid  is  a  quadrilat- 
eral having  only  two  of  its  op- 
posite sides  parallel. 


4.  A  diagonal  of  a  quadrilateral  or  other  polygon 
is  a  straight  line  joining  two  angles  not  adjacent  to 
each  other. 

5.  The  area  of  any  plane  figure  is  the  amount  of 
surface  which  it  contains. 

6.  Two  plane  figures  are  said  to  be  equivalent  when 
their  areas  are  equal. 


THEOREM  XIII. 

The  opposite,  sides  and  angles  of  a  parallelogram  are 
equal  to  each  other. 

Let  ABCD  be  a  parallelogram.     Dj — ^C 

It   is   to  be  proved  that  any  one 
of  its  sides  is  equal   to  the    side 
opposite,  and  any  one  of  its  an-    A 
gles  to  the  angle  opposite. 

Draw  the  diagonal  AC.  Now,  since  AC  intersects 
the  two  parallels  AD,  BC,  the  alternate  angles  DAC, 
BCA,  are  equal  (Theo.  Ill) ;  also,  since  it  intersects 
the  parallels  AB,  DC,  the  alternate  angles  BAC,  DCA, 
are  equal.  Hence,  the  two  triangles  ABC,  ADC,  have 
two  angles  of  the  one  equal  to  two  angles  of  the 
other,  and   the    included  side  AC  common ;  they  are, 


28  GEOMETRY. 

therefore,  equal  (Theo.  XI);  and  the  side  AB  is  equal 
to  the  side  DO  (Cor.,  Theo.  VII),  and  the  side  AD 
to  the  side  BC,  and  the  angle  B  to  the  angle  D ; 
also,  since  the  angle  DAC  is  equal  to  the  angle 
BCA,  and  the  angle  BAC  to  the  angle  DCA,  it  fol- 
lows that  the  whole  angle  BAD  is  equal  to  the  whole 
angle  BCD. 

Therefore,  the  opposite  sides  and  angles,  etc. 

Cor.  1.  A  diagonal  of  a  parallelogram  divides  it 
into  two  equal  triangles. 

Cor.  2.  If  ABCD  be  a  rectangular  parallelogram, 
AD  and  BC  will  both  be  perpendicular  to  each  of  the 
parallels  AB,  DC  (Def.  2,  Sec.  VI).  Hence,  the  per- 
pendicular distance  between  two  parallels  is  everywhere 
the  same. 

Schol.  Any  side,  as  AB, 
of  a  parallelogram  may  be 
taken  as  its  base;  and  a  per- 
pendicular, as  CE,  let  fall  a^ 
from  any  point  in  the  oppo- 
site parallel  on  the  base,  or  the  base  produced,  is 
called  the  altitude  of  the  parallelogram. 


THEOREM    XIV. 

If  a  quadrilateral  has   two   of   its  sides  equal   and 
parallel.,  it  is  a  parallelogram. 


Let  the  quadrilateral  ABCD  j)v _£ 

have  the  sides  AB,  DC,  equal  ,-'''   \ 

and  parallel.     Then  will  it  be  \ 

a  parallelogram.  A^- — — ^B 


BOOK   I.  29 

Join  AC ;  then,  since  AC  intersects  the  two  paral- 
lels AB,  DC,  the  alternate  angles  BAC,  DCA,  are 
equal  (Theo.  III).  Now,  in  the  triangles  ABC,  ADC, 
we  have  the  side  AB  equal  to  the  side  DC,  and  the 
side  AC  common,  and  the  included  angle  BAC  equal 
to  the  included  angle  DCA;  therefore,  the  two  tri- 
angles are  equal  (Theo.  VII),  and  the  angle  BCA  is 
equal  to  the  angle  CAD  (Cor.,  Theo.  VII),  and  these 
being  alternate  angles,  it  follows  that  AD  and  BC  are 
parallel  (Cor.  1,  Theo.  III).  Hence,  ABCD  has  its 
opposite  sides  parallel,  and  is  a  parallelogram  (Def. 
2,  Sec.  VI). 

Therefore,  if  a  quadrilateral,  etc. 

Sehol.  If  a  quadrilateral  has  its  opposite  sides  equal 
it  is  a  parallelogram.  For  the  triangles  ABC,  ADC, 
will  in  that  case  be  mutually  equilateral,  and  there- 
fore equal  (Cor.,  Theo.  XII) ;  and  the  parallelism  of 
the  opposite  sides  may  be  shown  by  the  equality  of 
the  alternate  angles,  as  above. 


THEOREM   XV. 
The  diagonals  of  a  parallelogram  bisect  each  other. 

Let  ABCD  be    a  parallelo-        ^ 
gram.     It  is  to  be  proved  that 
its    diagonals,  AC,  BD,  bisect                    ''Br- 
each other  in   E   (Def.  6,  Sec.  A^ ^b 

IV). 

In  the  triangles  AEB,  DEC,  since  the  angle  ABD 
is  equal  to  its  alternate  angle  BDC  (Theo.  Ill),  and 
the  angle  BAC  equal  to  its  alternate  angle  ACD,  and 
the  included  side  AB  equal  to  the  included  side  DC 


30 


GEOMETRY. 


(Theo.  XIII),  it   follows    that   the   two    triangles    are 
equal  (Theo.  XI) ;  consequently,  the  side  AE  is  equal 
to  the  side  EC,  and  BE  to  ED  (Cor.,  Theo.  VII)  ;  that 
is,  the  two  diagonals  bisect  each  other  in  E. 
Therefore,  the  diagonals,  etc. 


THEOREM    XVI. 


The  area  of  a  rectangle  is   equal   to   the   product  of 
its  base  by  its  altitude. 


Let  ABCD  be  a  rectangle. 
It  is  to  be  proved  that  its 
area  is  equal  to  the  product 
of  its  base  AB  by  its  altitude 
AD. 


D 

9 

f 


i 1 

e\ 


Let  AB    be    divided   into  a     A 
certain  number  of  equal  parts 

Ab,  be,  etc.,  taken  as  the  units  of  length ;  also,  let  AD 
be  divided  into  a  certain  number  of  the  same  units, 
A/,  fg,  etc.  From  b,  c,  etc.,  draw  straight  lines  par- 
allel to  AD,  and  from  /,  etc.,  draw  straight  lines 
parallel  to  AB.  Now,  it  is  evident  that  the  whole 
rectangle  is  divided  into  small  squares  (Cor.  2,  Theo. 
XIII),  each  equal  to  Abef;  which  may  be  taken  as 
the  unit  of  area.  Of  these  equal  squares  there  are  as 
many  in  the  tier  next  to  AB  as  there  are  units  of 
length  in  AB ;  and  there  are  as  many  equal  tiers  in 
the  whole  figure  as  there  are  units  of  length  in  AD. 
Therefore,  the  whole  number  of  square  units  in 
ABCD  is  equal  to  the  number  of  linear  units  in  AB 
multiplied  by  the  number  of  linear  units  in  AD. 

Hence,  the  area  of  a  rectangle,  etc. 


BOOK    I 


31 


Scliol.  If  AB  and  AD  are  incommensurable,  that 
is,  if  no  unit  can  be  found  into  which  they  can  both 
be  divided  without  leaving  a  remainder  in  one  of 
them,  the  theorem  will  still  hold  true ;  for  if  the  unit 
be  taken  smaller  and  smaller,  the  remainder  can  be 
made  less  than  any  assignable  quantity. 

When  the  linear  unit  is  one  inch,  the  unit  of  area 
is  a  square  inch ;  when  the  linear  unit  is  one  foot,  the 
unit  of  area  is  a  square  foot,  etc. 

Cor.  Since  the  base  and  altitude  of  a  square  are 
equal  (Def.  2,  Sec.  VI),  its  area  may  be  found  by 
multiplying  one  side  into  itself. 


THEOKEM    XVII. 

The  area  of  any  parallelogram  is  equal  to  the  area 
of  a  rectangle  having  tlie  same  base  and  altitude. 

Let  ABCD 
be  a  rectangle, 
and  ABFE 
any   parallelo- 


gram, 


on    the 


I) 

EC          F 

1  / 

/ 

/ 

/ 

/ 

B 


same  base  AB,  and  of  the  same  altitude,  namely 
the  perpendicular  distance  between  the  parallels  AB, 
DF.     Then  will  ABFE  be  equivalent  to  ABCD. 

Since  AB  and  DC  are  opposite  sides  of  a  parallelo- 
gram they  are  equal  (Theo.  XIII);  and  for  the  same 
reason  AB  and  EF  are  equal;  therefore  DC  is  equal 
to  EF  (Ax.  3).  Taking  away  each  of  these  in  turn 
from  the  whole  line  DF,  we  have  the  remainder  DE 
equal  to  the  remainder  CF.  But  DA  is  equal  to  CB 
(Theo.  XIII),  and  the  included  angle  ADE  is  equal  to 


32  GEOMETRY. 

the  included  angle  BCF  (Theo.  Ill) ;  therefore,  the 
triangles  ADE,  BCF,  are  equal  (Theo.  VII) ;  and 
hence  if  each  of  them  be  taken  away  in  turn  from  the 
whole  figure  ABFD,  the  remainder  ABFE  will  be 
equivalent  to  the  remainder  ABCD. 

Therefore,  the  area  of  any  parallelogram,  etc. 

Cor.  1.     The  area  of  any  parallelogram  is  equal  to 
the  product  of  its  base  by  its  altitude. 

Cor.  2.     Since  any  triangle,  as         D C 

ABC,  is  half   of   a  parallelogram 

ABCD    (Cor.   1,  Theo.   XIII)   on 

the  same  base  AB  and  having  the 

same    altitude    CE    (Def.    4,    Sec. 

V),  it   follows   that   the   area  of  a  triangle  is  equal  to 

half  the  product  of  its  base  by  its  altitude. 

THEOREM     XVIII. 

The  area  of  a  trapezoid  is  equal  to  half  the  product 
of  the  sum  of  its  parallel  sides  by  its  altitude. 

Let   ABCD  be   a  trapezoid  of         D o 

which  AB  and  DC  are  the   par-         /  y'    \ 

allel    sides.     Then    will    its    area        /      /''  \ 

be  equal  to   half  the    product  of      Ll, \ 

the  sum  of  AB   and  DC  into  its 

altitude,  namely,  the   perpendicular    distance    between 

AB  and  DC. 

Draw  the  diagonal  AC.  Now,  the  area  of  the 
triangle  ABC  (Cor.  2,  Theo.  XVII)  is  equal  to  half 
the  product  of  its  base  AB  into  its  altitude,  which  is 
the  same  as  the  altitude  of  the  trapezoid ;  again,  the 
area  of  the  triangle  ADC  is  equal  to  half  the  product 
of  its  base  DC  into  its  altitude,  which  is  also  the  same 


BOOK    I. 


33 


as   the   altitude    of  the   trapezoid.     Therefore,  adding 
together  the  area  of   the   two  triangles,  we    have   the 
area  of  the  whole  figure  equal  to  half  the  product  of 
the  sum  of  AB  and  DC  into  the  altitude. 
Hence,  the  area  of  a  trapezoid,  etc. 


THEOEEM    XIX. 

The  square  described  on  the  hypotenuse  of  a  right- 
angled  triangle  is  equivalent  to  the  sum  of  the  squares 
described  on  the  other  two  sides. 

Let  ABC  be  a  triangle 
right-angled  at  B.  It  is  to 
be  proved  that  the  square 
AEDC  is  equivalent  to  the 
sum  of  the  squares  ABGF 
and  BHIC. 

Join  FC,  BE,  and  draw 
BK  parallel  to  AE.  Also 
observe  that  since  ABC  and 

ABG  are  right  angles,  BC  and  BG  form  one  straight 
line. 

Now,  in  the  triangles  EAB,  CAF,  the  side  EA  is 
equal  to  the  side  CA,  since  they  are  sides  of  the  same 
square,  and  for  the  same  reason  the  side  AB  is  equal 
to  the  side  AF.  But  the  included  angles  EAB,  CAF, 
are  also  equal ;  for  each  of  them  is  composed  of  a 
right  angle  and  the  angle  CAB.  Therefore,  the  two 
triangles  are  equal  (Theo.  VII). 

But  since  the  triangle  CAF  and  the  square  BAFG 
have  the  same  base  AF,  and  the  same  altitude,  namely 
the  perpendicular  distance  between  the  parallels  AF, 


34  GEOMETRY. 

CG,  the  square  must  be  double  of  the  triangle  (Cors.  1 
and  2,  Theo.  XVII).  For  like  reason  the  parallelo- 
gram AEKL  is  double  of  the  triangle  EAB.  But 
doubles  of  equals  are  equals ;  therefore,  the  paral- 
lelogram AEKL  is  equivalent  to  the  square  BAFG. 
In  the  same  manner  (by  joining  AI  and  BD),  it 
may  be  shown  that  the  parallelogram  CDKL  is  equiv- 
alent to  the  square  BCIH.  Hence,  the  whole  square 
ACDE  is  equivalent  to  the  sum  of  the  squares 
BAFG,  BCIH. 

Therefore,  the  square  described,  etc. 

Cor.  1.  The  hypotenuse  is  equal  to  the  square 
root  of  the  sum  of  the  squares  of  the  other  two  sides. 

Cor.  2.  The  square  of  either  of  the  sides  contain- 
ing the  right  angle  is  equivalent  to  the  square  of  the 
hypotenuse  diminished  by  the  square  of  the  other 
side.  By  taking  the  square  root  of  the  remainder  the 
side  itself  will  be  found. 

Cor.  3.  If  two  right-angled  triangles  have  the  hy- 
potenuse and  one  side  of  the  one  respectively  equal 
to  the  hypotenuse  and  one  side  of  the  other,  the 
third  sides  will  also  be  equal. 

EXEHCISES. 

1.  If  the  side  of  a  square  be  36  inches,  what  is  its 
area  in  square  inches  ?     What  in  square  feet  ? 

2.  If  the  base  of  a  parallelogram  be  3  feet  and  its 
altitude  4  feet  and  6  inches,  what  is  its  area? 

3.  If  the  base  of  a  triangle  be  50  yards  and  its 
altitude  20  yards,  what  is  its  area? 

4.  If  the  parallel  sides  of  a  trapezoid   be  12  rods 


BOOK    I.  35 

and    16    rods,  and   its   altitude    8  J   rods,   what   is   its 
area  ? 

5.  If  the  sides  containing  the  right  angle  of  a  riffht- 
angled  triangle  be  3  and  4,  what  is  the  length  of  the 
hypotenuse  ? 

6.  If  the  hypotenuse  be  10  and  one  of  the  sides  8, 
what  is  the  length  of  the  other  side  ? 

7.  Prove  that  if  a  parallelogram  has  one  right 
angle  it  is  a  rectangle. 

8.  Prove  that  the  diagonals  of  a  rectangle  are  equal 
to  each  other. 

9.  Prove  that  a  perpendicular  is  the  shortest  line 
that  can  be  drawn  to  a  straight  line  from  a  point 
without  it ;  also,  that  of  two  oblique  lines  that  which 
is  furthest  from  the  perpendicular  is  the  longest 
(Theo.  XIX). 


SEC.  VII.— OF   POLYGONS   IN    GENERAL. 
DEFINITIONS. 

1.  A  polygon  of  five  sides  is  called  a  pentagon  ; 
one  of  six  sides,  a  hexagon  ;  of  seven  sides,  a  hepta- 
gon ;  of  eight  sides,  an  octagon  ;  of  ten  sides,  a  deca- 
gon ;  of  twelve  sides,  a  dodecagon. 

2.  A  regular  polygon  is  one  which  is  both  equilat- 
eral and  equiangular.  Squares  and  equilateral  tri- 
angles are  species  of  regular  polygons. 

8.  The  perimeter  of  a  polygon  is  the  sum  of  all  its 

sides. 


36 


GEOMETRY. 


THEOREM    XX. 

The  sum  of  all  the  angles  of  any  polygon  is  equal  to 
twice  as  many  rigid  angles,  wanting  four,  as  the  figure 
has  sides. 

Let  ABODE  be  any  polygon. 
It  is  to  be  proved  that  the  sum 
of  all  its  angles  A,  B,  C,  D,  E, 
is  equal  to  twice  as  many  right 
angles,  wanting  four,  as  it  has 
sides. 

From  the  vertex  of  each  angle 
draw  straight  lines  AF,  BF,  etc.,  to  any  point  F 
within,  thus  dividing  the  polygon  into  as  many  tri- 
angles as  it  has  sides.  Now,  since  the  angles  of  each 
triangle  are  together  equal  to  two  right  angles  (Theo. 
V),  it  follows  that  the  angles  of  all  the  triangles  are 
together  equal  to  twice  as  many  right  angles  as  the 
figure  has  sides.  But  the  sum  of  the  angles  A,  B,  C, 
D,  E,  is  equal  to  the  sum  of  the  angles  of  all  the 
triangles,  wanting  the  angles  about  the  point  F ;  and 
the  angles  about  the  point  F  are  together  equal  to 
four  right  angles  (Cor.  2,  Theo.  I).  Therefore,  the  sum 
of  the  angles  A,  B,  C,  D,  E,  is  equal  to  twice  as 
many  right  angles,  wanting  four,  as  the  polygon  has 
sides. 

That  is,  the  sum  of  all  the  angles  of  any  poly- 
gon, etc. 

Schol.  The  area  of  a  polygon  may  be  found  by 
dividing  it  into  triangles  as  above  (or  by  diagonals 
drawn  from  any  angle  A  to  the  opposite  angles),  then 
finding  the  areas  of  these  triangles  separately  and 
adding  them  together. 


BOOK    I.  37 

THEOEEM    XXI. 

A   regular  polygon  may   he   divided    into   as   many 
equal  triangles  as  it  has  sides. 


m    E 


Let  ABCDEF  be  a  regular 
polygon.  Draw  AG  and  BG 
bisecting  the  angles  A  and  B ; 

also,  draw  GC,  GD,  etc.,  to  the    A/ _-V         ._^D 

other  angles. 

Now,  since  the  triangles 
AGB,  BGC,  have  the  side  AB  B  c 

equal  to  the  side  BC  (Def.  2,  Sec.  VII),  and  the  side 
BG  common,  and  the  included  angle  ABG  equal  to 
the  included  angle  CBG,  the  two  triangles  are  equal 
to  each  other  (Theo.  VII).  Hence  the  angle  GAB  is 
equal  to  the  angle  GCB  (Cor.,  Theo.  VII).  But  GAB 
is  by  construction  half  of  BAF ;  therefore,  GCB  must 
be  half  of  the  equal  angle  BCD,  that  is,  BCD  is  bi- 
sected by  GC.  Hence,  it  may  be  shown  in  the  same 
manner  as  before  that  the  triangles  BGC,  CGD,  are 
equal :  and  so  on  of  the  other  triangles  in  succes- 
sion ;  of  which  there  is  one  on  each  side  of  the 
polygon. 

Therefore,  a  regular  polygon,  etc. 

ScJwl.  The  center  of  a  regular  polygon  is  the  point 
G,  found  by  bisecting  any  two  of  its  angles.  A  per- 
pendicular, as  Gm,  let  fall  from  the  center  upon  any 
side  of  a  regular  polygon  is  called  its  apothegm.  It 
is  the  altitude  of  the  triangle  EFG,  and  consequently 
of  each  of  the  equal  triangles  into  which  the  polygon 
is  divided. 

Cor.     Since  the  area  of  each  triangle  is   equal  to 


38 


GEOMETRY. 


half  the  product  of  its  base  by  its  altitude,  the  area 
of  all  the  triangles  taken  together  is  equal  to  half 
the  product  of  the  sum  of  their  bases  by  their  com- 
mon altitude :  that  is,  the  area  of  a  regular  polygon 
is  equal  to  half  the  product  of  its  perimeter  by  its 
apothegm. 

EXEECISES. 

1.  To  how  many  right  angles  are  all  the  angles  of 
a  quadrilateral  equal  ?  Of  a  pentagon  ?  Of  a  hexa- 
gon ?     Of  a  heptagon  ?  etc. 

2.  What  is  the  area  of  a  regular  pentagon  whose 
side  is  25,  and  the  apothegm  17.205? 

3.  Prove  that  each  angle  of  a  regular  hexagon  is 
equal  to  four  thirds  of  a  right  angle. 


SEC.  VIII.— OF   THE    CIRCLE. 
DEFINITIONS. 

1.  A  circle  is  a  plane  figure  bounded  by  a  curve, 
called  the  circumference,  every  point  in  "which  is 
equally  distant  from  a  point  within,  called  the  CENTER. 

2.  A  straight  line  drawn  from 
the  center  to  any  point  of  the  cir- 
cumference is  called  the  RADIUS. 

A  double  radius,  that  is,  a 
straight  line  passing  through  the 
center,  and  terminated  both  ways 
by  the   circumference,    is    called 

the  DIAMETER. 

Cor.  In  the  same  circle,  all  radii  are  equal ;  also, 
all  diameters. 


BOOK    I 


39 


3.  A  TANGENT  is  a  straight 
line  which  meets  the  circum- 
ference, but  being  produced 
does  not  cut  it. 

4.  Any  portion  of  the  cir- 
cumference is  called  an  arc. 
The  chord  of  an  arc  is  the 
straight  line  joining  its  ex- 
tremities. A  chord  produced  one  way  beyond  the 
circumference  is  called  a  secant. 

5.  A  segment  of  a  circle 
is  the  part  contained  by  an 
arc  and  its  chord. 

A  sector  is  the  part  con- 
tained by  an  arc  and  two 
radii  drawn  to  its  extremities. 

6.  Half  a  circle  is  called  a 


semicircle;     half    a    circum- 


ference, a  semi-circumference.     A  quarter  of  a  circle 
or  of  a  circumference  is  called  a  quadrant. 

7.  An  inscribed  angle  is  one  which  has  its  vertex 
in  tire*  circumference,  and  is  contained  by  two  chords. 

A  polygon  is  said  to  be  inscribed  in  a  circle  when 
all  its  sides  are  chords  of  that  circle. 

A  circle  is  said  to  be  inscribed  in  a  polygon  when 
all  the  sides  of  the  polygon  are  tangents  to  the  circle. 

When  one  figure  is  inscribed  in  another,  the  latter 
is  said'  to  be  circumscribed  about  the  former. 


THEOREM"    XXII. 

A  diameter  divides  the   circle   and  its   circumference 
into  two  equal  parts. 


40 


GEOMETRY 


Let  ABCD  be  a  circle  of 
which  E  is  the  center  and 
AC  a  diameter.  It  may  be 
shown  that  the  segment  ABC  A 
is  equal  to  the  segment  ADC, 
and  the  arc  ABC  to  the  arc 
ADC. 

Let  the  segment  ABC  be 
applied  to  the  segment  ADC,  the  line  AC  remaining 
common  ;  then  the  arc  ABC  will  exactly  coincide  with 
the  arc  ADC :  for  if  not,  suppose  some  point  in  ABC 
to  fall  within  or  without  ADC ;  then  we  shall  have 
points  in  the  circumference  unequally  distant  from 
the  center,  which  is  contrary  to  the  definition  of  a 
circle.  Hence,  the  arc  ABC  is  equal  to  the  arc  ADC, 
and  the  segment  ABC  to  the  segment  ADC  (Ax.  11). 

Therefore,  a  diameter  divides,  etc. 

Schol.  In  a  similar  manner  it  may  be  proved  that 
two  circles  of  equal  diameters  or  radii  are  equal  to 
each  other. 


THEOREM    XXIII. 

If  a  straight  line  is  perpendicular  to  a  diameter  at 
its  extremity,  it  is  a  tangent  to  the  circle. 

Let  the  straight  line  EF 
be  perpendicular  to  the  di- 
ameter AC  at  its  extremity 
C.  Then  will  EF  be  a  tan- 
gent to  the  circle  ABCD. 

In  EF  take  any  two  points 
m  and  n,  one  on  each  side 
of    C,    and    connect   them   by 


BOOK    I, 


41 


straight  lines  with  the  center  G.  Now,  since  GCm  is 
a  right-angled  triangle,  the  square  of  Gm  is  greater 
than  the  square  of  GC  (Theo.  XIX) ;  hence,  Gm  is 
greater  than  GC,  that  is,  greater  than  the  radius. 
Therefore,  the  point  m,  however  near  it  may  be 
to  C,  is  necessarily  without  the  circle.  The  same 
may  be  proved,  in  like  manner,  of  the  point  n,  or 
any  other  point  in  EF  on  either  side  of  C.  Hence, 
EF  can  meet  the  circumference  only  in  the  point 
C,  and  it  is  consequently  a  tangent  (Def.  3,  Sec. 
VIII). 

Therefore,  if  a  straight  line,  etc. 

Cor.     If  a  chord  be  drawn  perpendicular  to  a  tan- 
gent at  the  point  of  contact,  it  will  be  a  diameter. 


THEOEEM   XXIV. 

Equal  arcs  in  a  circle  have  equal   chords,  and  sub- 
tend equal  angles  at  the  center. 

Let  AD  and  BC  be  equal 
arcs,  and  E  the  center  of  the 
circle.  It  is  to  be  proved 
that  the  chords  AD,  BC  are 
equal ;  also,  the  subtended 
angles  AED,  BEC. 

Let  the  sector  EBC  be  ap- 
plied to  the  sector  EAD,  so  that  the  line  EB  shall  fall 
upon  the  line  EA,  the  point  E  remaining  common  ; 
then,  since  these  lines  are  equal,  being  radii  (Cor., 
Def.  2,  Sec.  8),  the  point  B  will  fall  upon  the  point 
A ;  and,  since  all  points  in  the  circumference  are 
equally  distant  from  the  center,  the  arc  BC  will  fall 
Evans    Geometry. — 4 


42  GEOMETRY. 

upon  the  arc  AD;  and  these  arcs  being  by  hypothesis 
equal,  the  point  C  will  fall  upon  the  point  D ;  there- 
fore (Ax.  10),  the  chord  BC  will  coincide  with  the 
chord  AD,  and  be  equal  to  it. 

Again,  in  the  triangles  EAD,  EBC,  the  two  sides 
EA,  ED,  are  equal  to  the  two  sides  EB,  EG,  since 
they  are  all  radii;  and  the  base  AD  has  just  been 
proved  equal  to  the  base  BC ;  therefore,  the  two 
triangles  are  mutually  equiangular  (Theo.  XII),  and 
the  angle  AED  is  equal  to  the  angle  BEC  (Cor., 
Theo.  VII). 

Hence,  equal  arcs  in  a  circle,  etc. 

Sclwl.  If  the  whole  circumference  be  divided  into 
any  number  of  equal  arcs,  and  radii  be  drawn  to  all 
the  points  of  division,  the  whole  compass  about  the 
center  will  be  divided  into  the  same  number  of  equal 
angles.  Hence,  an  arc  may  be  taken  as  the  measure 
of  its  subtended  angle.  To  this  end  the  circumference 
is  conceived  to  be  divided  into  360  equal  parts,  called 
degrees,  each  degree  into  60  minutes,  and  each  minute 
into  60  seconds.  Then  whatever  number  of  degrees, 
etc.,  an  arc  contains,  the  same  number  will  denote 
the  magnitude  of  the  angle  at  the  center  which  it 
subtends. 

Cor.  A  quadrant  is  an  arc  of  90  degrees  ;  and  a 
right  angle,  being  measured  by  a  quadrant,  is  an  angle 
of  90  degrees. 


THEOREM    XXV. 

If  a  radius   is  perpendicular   to   a   chord,  it  bisects 
the  chord,  and  also  the  arc  which  the  chord  subte?ids. 


BOOK    I 


43 


Let  the  radius  DB  be  perpen- 
dicular to  the  chord  AC.  It  is 
to  be  proved  that  it  bisects  AC, 
and  also  the  arc  ABC. 

Draw  the  radii  DA,  DC. 
Now,  since  the  right-angled  tri- 
angles AED,  CED,  have  the 
hypotenuse    AD    equal    to    the 

hypotenuse  CD,  and  the  side  ED  common,  it  follows 
that  the  third  side  AE  is  equal  to  the  third  side  CE 
(Cor.  3,  Theo.  XIX) ;  that  is,  AC  is  bisected  in  E. 

Again,  since  the  triangles  AED,  CED,  are  mutually 
equilateral,  the  angle  ADE  is  equal  to  the  angle  CDE 
(Theo.  XII).  Hence,  the  arc  AB  which  measures  the 
former  angle  is  equal  to  the  arc  CB  which  measures 
the  latter  (Schol.,  Theo.  XXIV). 

Therefore,  if  a  radius  is  perpendicular,  etc. 


THEOKEM    XXVI. 

An  inscribed  angle  is  measured  by  half  the  arc  on 
which  it  stands. 


Let  ACB  be  an  angle  in- 
scribed in  a  circle.  It  is  meas- 
ured by  half  the  arc  AEB. 


First, 


the 


D 


,  suppose  the  center 
to  be  within  the  angle.  Draw 
the  diameter  CE  ;  also,  join  DA 
and  DB.  Now,  since  AD  and 
DC  are  radii,  the  triangle  ADC  is  isosceles ;  hence, 
the  angles  DAC,  DCA^are  equal  (Theo.  VIII).  But 
the  exterior  angle  ADE  is  equal  to  the  sum  of  the 
two    opposite    interior    angles    DAC,    DCA    (Cor.    1, 


44 


GEOMETRY. 


Theo.  V) ;  it  is  consequently  double  of  DCA ;  and 
since  ADE  is  measured  by  the  arc  AE  (Scbol.,  Theo. 
XXIV),  it  follows  that  DCA  is  measured  by  half  of 
AE.  In  the  same  manner  it  may  be  shown  that 
DCB  is  measured  by  half  of  BE.  Therefore,  the 
whole  angle  ACB  is  measured  by  half  of  the  whole 
arc  AEB. 

Next,  let  the  center  D  be 
without  the  angle  ACB.  By 
the  above  demonstration,  the 
angle  ACE  is  measured  by  half 
the     arc    AE,    and    the     anfr^ 


angle 


BCE   is   measured  by   half  the 

arc  BE ;  therefore,  BCA,  which 

is  the  difference  of  these  two  angles,  is  measured  by 

half  the    difference    of  the   two    arcs,  that   is,  by  half 

of  AB. 

Hence,  an  inscribed  ancrle,  etc. 


Cor.  1.  Angles  inscribed  in 
the  same  segment,  as  AEC  and 
ADC,  are  equal;  for  they  are 
measured  by  half  the  same  arc 
ABC. 


Cor.  2.  Any  angle  inscribed  in  a  semicircle  is  a 
right  angle;  being  measured  by  half  a  semi-circum- 
ference. 


THEOEEM    XXVII. 


An   angle    contained    bg    a   tangent   and  a  chord  is 
measured  by  half  the  intercepted  arc. 


BOOK    1. 


45 


A 


B 


Let  AC  be  a  tangent  to  a 
circle,  and  BF  a  chord  drawn 
from  the  point  of  contact  B. 
Then  will  the  angle  ABF  be 
measured  by  half  the  arc  BF, 
and  the  angle  CBF  by  half  the 
arc  BDF. 

Draw  BE  perpendicular  to 
AC,  and  it  will  be  a  diameter  of  the  circle  (Cor., 
Theo.  XXIII).  Now,  since  ABE  is  a  right  angle,  it  is 
measured  by  half  the  semi-circumference,  BFE  (Cor., 
Theo.  XXIV) ;  and  since  the  part  EBF  is  an  inscribed 
angle,  it  is  measured  by  half  the  arc  FE  (Theo. 
XXVI) ;  therefore,  the  remaining  angle  ABF  is  meas- 
ured by  half  the  remaining  arc  BF. 

Again,  because  CBE  is  measured  by  half  the  semi- 
circumference  BDE,  and  EBF  by  half  the  arc  EF,  it 
follows  that  the  whole  angle  CBF  is  measured  by 
half  the  whole  arc  BDF. 

Therefore   an  angle  contained,  etc. 


THEOREM    XXVIII. 

If  the   circumference  be  divided  into  any  number  of 
equal  arcs,  their  chords  will  form  a  regular  polygon. 

Let  the  arcs  AB,  BC,  etc., 
be  all  equal  to  each  other. 
Then  will  the  inscribed  polygon 
formed  by  their  chords  AB,  BC, 
etc.,  be  a  regular  polygon. 

Draw  the  radii  AG,  BG,  CG, 

etc.     Now,    since   two    sides    in 


46  GEOMETRY. 

each  of  the  triangles  AGB,  BGC,  etc.,  are  radii ;  and 
since  their  bases  AB,  BC,  etc.,  are  equal,  being  chords 
of  equal  arcs  (Theo.  XXIV) ;  it  follows  that  these 
triangles  are  mutually  equiangular  (Theo.  XII).  But 
they  are  also  isosceles ;  hence  (Theo.  VIII),  all  the 
angles  at  their  bases,  GBA,  GBC,  GCB,  GCD,  etc, 
are  equal ;  hence,  also,  the  sums,  ABC,  BCD,  etc., 
that  is,  all  the  angles  of  the  polygon,  are  equal.  And 
its  sides  have  already  been  shown  to  be  equal.  It  is, 
therefore,  a  regular  polygon  (Def.  2,  Sec.  VII). 

Hence,  if  the  circumference,  etc. 

Cor.  1.  If  the  polygon  ABCDEF  be  a  hexagon, 
the  angle  AGB,  being  measured  by  one-third  of  the 
semi-circumference,  will  be  one-third  of  two  right 
angles  (Cor.,  Theo.  XXIV).  Hence,  either  of  the  two 
equal  angles  at  the  base  of  the  triangle,  as  GBA,  will 
also  be  one-third  of  two  right  angles  (Theo.  V). 
Therefore,  AB  is  equal  to  AG  (Cor.,  Theo.  IX) ;  that 
is,  the  side  of  a  regular  hexagon  is  equal  to  the  radius 
of  the.  circumscribed  circle. 

Cor.  2.  A  perpendicular,  as  Gm,  let  fall  from  the 
center  of  the  circumscribed  circle  on  any  side  of  a 
regular  polygon,  will  be  the  apothegm  of  the  polygon 
(Schol.,  Theo.  XXI). 

Cor.  3.  If  from  the  center  G,  with  the  apothegm 
Gm  as  radius,  a  circle  be  drawn,  the  side  FE,  and 
consequently  all  the  other  sides  of  the  regular  poly- 
gon, will  be  tangents  to  that  circle  (Theo.  XXIII), 
and  the  circle  will  be  inscribed  in  the  polygon  (Def. 
7,  Sec.  VIII). 


BOOK    I 


47 


THEOREM    XXIX. 

The  area  of  a  circle  is  equal  to  half  the  product  of 
its  radius  by  its  circumference. 

Let    ABC    be     a    circle,    of 
which  DB  or  DC  is  radius. 

If  we  conceive  the  whole  cir- 
cumference to  be  divided  into 
equal  arcs  so  small  that  the 
points  of  division  may  be  re- 
garded as  infinitely  near  to  each 
other,  the  perimeter  of  the  reg- 
ular polygon  formed  by  their  chords  (Theo.  XXVIII) 
will  coincide  with  the  circumference,  and  the  area  of 
the  polygon  will  be  equal  to  the  area  of  the  circle  ; 
also,  the  apothegm  of  the  polygon  (Cor.  2,  Theo. 
XXVIII)  will  be  equal  to  the  radius  of  the  circle. 
But  the  area  of  the  polygon  will  be  equal  to  half  the 
product  of  its  apothegm  by  its  perimeter  (Cor.,  Theo. 
XXI).  Therefore,  the  area  of  the  circle  must  be 
equal  to  half  the  product  of  its  radius  by  its  circum- 
ference. 

Cor.  1.  In  like  manner  it  may  be  shown  that  the 
area  of  any  sector,  as  BECD,  is  equal  to  half  the 
product  of  the  radius  into  its  arc. 

Cor.  2.  The  area  of  a  segment  BEC,  less  than  a 
semicircle,  may  be  found  by  subtracting  the  triangle 
BDC  from  the  sector  BECD.  The  area  of  a  segment 
BAC,  greater  than  a  semicircle,  may  be  found  by 
adding  the  triangle  BDC  to  the  sector  BACD. 


48 


GEOMETRY. 


THEOREM    XXX. 

The  radius,  and  one  side  of  a  regular  inscribed 
polygon  Icing  given,  the  side  of  a  regular  inscribed 
polygon  of  twice  the  number  of  sides  mag  be  found. 

Let  AB  be  the  side  of  a  reg- 
ular inscribed  polygon  in  a 
circle  whose  radius  is  CA  or 
CB.  DraAV  CE  perpendicular  to 
AB.  This  will  bisect  AB  and 
also  the  arc  AEB  (Theo.  XXV). 
Hence,  the  chords  AE,  EB 
(Theo.  XXVIII),   will    be   sides 

of  a  regular  inscribed  polygon  having  twice  as  many 
sides  as  the  first. 

Now,  if  the  lengths  of  CB  and  AB  are  given,  we 
shall  have,  in  the  right-angled  triangle  CDB,  the  hy- 
potenuse CB  and  the  side  BD;  hence,  we  may  find 
the  other  side  DC  (Cor.  2,  Theo.  XIX).  Subtracting 
this  from  the  radius  CE,  we  shall  have  DE.  Then,  in 
the  right-angled  triangle  BDE,  we  shall  have  the  two 
sides  BD,  DE,  from  which  we  can  find  the  hypotenuse 
EB  (Cor.  1,  Theo.  XIX) ;  which  is  the  side  required. 

Schol.  If  the  diameter  be  1,  the  side  of  a  regular 
inscribed  hexagon  will  be  \  (Cor.  1,  Theo.  XXVIII). 
From  this  we  can  find  the  side,  and  consequently  the 
perimeter,  of  a  regular  inscribed  dodecagon;  from  that 
Ave  can  find  the  perimeter  of  a  regular  inscribed  polygon 
of  24  sides,  etc.  By  carrying  this  calculation  on,  to 
polygons  of  an  indefinitely  large  number  of  sides,  it 
is  found  that  the  perimeter,  though  it  increases  at 
every  step,  never  exceeds  3.14159,  except  by  decimal 
figures  beyond  lkjfr  9;  that  is,  beyond  the  fifth  place 
of    decimals.     Mcnce,  since    the    perimeter  ultimately 


M.cnc 


BOOK    I.  49 

coincides  with  the  circumference,  it  follows  that  the 
circumference  can  not  differ  from  8.14159,  except  by 
decimals  beyond  the  fifth  place.  Disregarding  these, 
as  of  trifling  value,  we  may  conclude  that  the  circum- 
ference of  a  circle  whose  diameter  is  1,  is  3.14159. 

Cor.     If  the    diameter  is    1,   the    area   is    equal  to 

1  yQ  141  ^Q 

"X     "        (Theo.  XXIX),  which  by  reduction  is  .7854. 

EXEECISES. 

1.  How  many  degrees  are  contained  in  an  arc  cut 
off  by  the  side  of  an  inscribed  square  ?  Of  a  regular 
inscribed  pentagon  ?     Hexagon  ?     Heptagon  ?  etc. 

2.  What  is  the  area  of  a  sector  whose  arc  is  25 
feet,  in  a  circle  whose  diameter  is  16  feet? 

3.  Prove  that  in  an  inscribed  quadrilateral  the  sum  of 
any  two  opposite  angles  is  180  degrees  (Theo.  XXYI). 

4.  Prove  that  of  two  unequal  chords  the  less  is  the 
further  from  the  center  (Theo.  XXV,  and  Cor.  2, 
Theo.  XIX). 

SEC.  IX.— PROBLEMS   IN   CONSTRUCTION. 

PROBLEM   I. 

To  bisect  a  given  straight  line. 

Solution.  Let  AB  be 
the  given  line.  From  A 
as  a  center,  with  a  radius 
obviously  greater  than 
one-half  of  AB,  describe 
the  arc  CFD.  From  B 
as  a  center,  with  the 
same  radius,  describe  the  arc  CED.  Join  the  points 
of  intersection  C  and  D. 
Evans    Geometry. — 5 


50 


GEOMETRY. 


Now,  since  the  opposite  sides  of  the  quadrilateral 
ACBD  are  equal,  it  is  a  parallelogram  (Schol.,  Theo. 
XIV).  But  the  diagonals  of  a  parallelogram  bisect 
each  other  (Theo.  XV).     Therefore,  CD  bisects  AB. 


PROBLEM  II. 

At  a  given  point  in  a  straight  line,  to  erect  a  perpen- 
dicular to  that  line. 


f 


A     D 


E     B 


Solution.  Let  AB  be 
an  indefinite  straight  line, 
and  C  the  given  point  in 
it.  Take  the  points  D 
and  E  equally  distant 
from  C.  From  D  as  a  center,  with  a  radius  greater 
than  DC,  describe  an  arc;  and  from  E  as  a  center, 
with  the  same  radius,  describe  another  arc  intersecting 
the  former  at  some  point  F.  Now,  draw  FC,  and  it 
will  be  the  perpendicular  required. 

Join  FD  and  FE.  Then,  since  the  triangles  DFC, 
EFC,  are  by  construction  mutually  equilateral,  the 
angles  FCD,  FCE,  are  equal  (Theo.  XII) ;  they  are, 
therefore,  right  angles,  and  CF  is  perpendicular  to 
AB  at  the  point  C  (Defs.  2  and  3,  Sec.  IV.) 


PROBLEM   III. 

To   draw  a  perpendicular  to  a  straight  line  from,  a 
given  point  without. 

Solution.  Let  C  be 
the  given  point  without 
the  straight  line,  AB. 
From  C  as  a  center,  with 
a  radius  greater  than  the 


BOOK    I. 


51 


shortest  distance  to  the  line,  draw  an  arc  cutting  AB 
in  D  and  E.  Bisect  DE  in  F  (Prob.  I).  Then  draw 
CF,  and  it  will  be  the  perpendicular  required. 

Join  CD  and  CE.  Now,  since  the  triangles  DCF, 
EOF,  are  mutually  equilateral,  the  angles  CFD,  CFE, 
are  equal  (Theo.  XII) ;  they  are,  therefore,  right 
angles,  and  CF  is  a  perpendicular  to  AB  from  the 
point  C. 

PEOBLEM   IV. 

Solution.  Let  BAC  be  the  given 
angle.  From  A  as  a  center,  with  any 
radius,  draw  an  arc  BDC  intersecting 
both  sides  of  the  angle.  Then  draw 
AD  perpendicular  to  the  chord  BC 
(Prob.  Ill),   and   it    will   bisect   the   angle. 

For,  since  the  perpendicular  AD  is  radius,  it  bisects 
the  arc  BDC  (Theo.  XXV).  Therefore,  the  arcs  BD, 
DC,  being  equal,  the  subtended  angles  BAD,  DAC, 
must  also  be  equal  (Theo.  XXIV),  and  the  angle  BAC 
is  bisected  by  AD. 


PEOBLEM  V. 

At  a  given  point  in  a  straight  line,  to  make  an  angle 
equal  to  a  given  angle. 

Solution.  Let  ACB  be 
the  given  angle,  and  F 
the  given  point  in  a 
straight  line  FE.     From     c  sf 

C,  as  a  center,  with  any  radius  as  CB,  describe  an  arc 
BA  intersecting  both  the  sides  of  the  angle ;  also, 
draw  the  chord  BA.     Then,  from  F  as  a  center,  with 


52  GEOMETRY. 

a  radius  FE  equal  to  CB,  draw  an  indefinite  arc ;  and 
from  the  center  E,  with  a  radius  equal  to  the  chord 
BA,  draw  an  arc  intersecting  the  other  at  D;  also, 
join  DE  and  DF.  The  angle  DFE  is  the  angle  re- 
quired. For,  by  equality  of  the  triangles  DEF,  ABC, 
the  angle  F  is  equal  to  the  angle  C  (Theo.  XII). 


PROBLEM    VI. 

Through  a  given  'point  to  draw  a  straight  line  par- 
allel to  a  given  line. 

Solution.     Let    AB  E _£_  -i 

be  the    given  straight 
line,  and  C  the  given 

point.     In    AB     take        

any  point  D,  and  join 

CD.  Through  C  draw  EF,  making  the  angle  ECD 
equal  to  the  angle  CDB  (Prob.  V).  These  being 
alternate  angles,  the  straight  line  EF  must  be  parallel 
to  AB  (Cor.  1,  Theo.  III). 

PROBLEM   VII. 

Two   angles   of  a   triangle    being  given,  to  find    the 
third. 

Solution.  At  any  point  C  in 
a  straight  line  AB,  make  ACD 
equal  to  one  of  the  given 
angles  (Prob.  V),  and  DCE 
equal   to   the   other;    then  will    • 

ECB  be  equal  to  the  third  angle.  For  the  sum  of 
the  three  angles  at  C  is  two  right  angles  (Cor.  1, 
Theo.  I) ;  which  is  also  the  sum  of  the  three  angles  of 
a  triangle  (Theo.  V). 


BOOK  I. 

PROBLEM   VIII. 

Given  the  hypotenuse  and  one  side  of  a  right-angled 
triangle,  to  construct  the  triangle. 

Solution.  Draw  AB  equal  to  c 
the  given  side.  At  B  erect  a 
perpendicular  BC  (Prob.  II). 
From  A  as  a  center,  with  a 
radius  equal  to "  the  given  hy- 
potenuse, describe  an  arc  inter- 
secting BC  at  D.  Join  AD. 
It    is    evident    that    ABD    is    the    triangle    required. 


PEOBLEM    IX. 

To  draw  a  circle  through  three  given  points. 

Solution.  Let  A,  B,  and  C 
be  the  given  points.  Draw 
the  straight  lines  AB  and  BC, 
and  from  their  middle  points 
erect  the  perpendiculars  DF 
and  EF.  Also,  join  AF,  BF, 
and  CF. 

Now,  because  the  triangles 
AFD,  BFD,  have  the  side  AD  equal  to  the  side  BD, 
and  the  side  DF  common,  and  the  included  angle 
ADF  equal  to  the  included  angle  BDF,  the  third  side 
AF  is  equal  to  the  third  side  BF  (Theo.  VII).  In 
the  same  manner  it  may  be  shown  that  BF  is  equal 
to  CF.  Therefore,  the  three  lines  AF,  BF,  and  CF 
are  equal ;  and  if  from  F  as  a  center,  with  one  of 
these  equals  as  radius,  a  circle  be  drawn,  it  will  pass 
through  the  three  points  A,  B,  and  C. 


54  GEOMETB  Y. 

Cor.  Hence,  the  center  of  a  given  circle  may  be 
found  by  erecting  perpendiculars  on  the  middle  points 
of  two  chords,  and  producing  them  till  they  meet. 


EXERCISES. 

1.  Given  the  three  sides  of  a  triangle,  to  construct 
"the  triangle. 

2.  Given  two  angles  and  the  iacluded  side,  to  con- 
struct the  triangle. 

3.  Given  two  adjacent  sides  and  the  included  angle 
of  a  parallelogram,  to  construct  the  parallelogram. 

4.  To  draw  a  diameter  of  a  given  circle. 

5.  At  a  given  point  in  the  circumference  of  a  circle, 
to  draw  a  tangent  to  the  circle. 


PLANE   GEOMETBY. 


BOOK   II. 


PROPORTIONS  OF  MAGNITUDES  LYING  IN  THE  SAME  PLANE. 


SEC.  X.— OF   POLYGONS. 


DEFINITIONS. 


1.  In  mutually  equiangular  figures,  the  sides  which 
are  similarly  situated  with  respect  to  the  equal  angles, 
are  called  homologous 
sides.     Thus, 

If  the  angles  A,  B,  C, 
are  respectively  equal 
to  the  angles  a,  b,  e, 
the  side  AB  is  homol- 
ogous to  the  side  ab,  the  side  BC  to  the  side  be,  and 
CA  to  ca. 

Two  triangles  or  other  polygons  are  called  similar 
when  they  are  mutually  equiangular,  and  their  homol- 
ogous sides  are  proportional. 

2.  Magnitudes  are  called  proportionals  when  the 
first  lias  the  same  ratio  to  the  second,  that  the  third 
has  to  the  fourth,  the  fifth  to  the  sixth,  etc. 

The  first  terms  of  the  several  equal  ratios  are  called 
the  antecedents,  and  the  second  the  consequents.     Thus, 

If  A  :  B  :  :  C  :  D  :  :  E  :  F,  the  terms  A,  C,  E,  are 
the  antecedents,  and  B,  D,  F,  the  consequents. 

(55) 


56  GEOMETRY. 

If  four  magnitudes  are  in  proportion,  the  first  and 
fourth  terms  are  called  the  extremes,  and  the  second 
and  third  the  means.  In  this  case  the  last  term  is 
called  a  fourth  projjortional  to  the  three  others. 

If  the  first  of  three  magnitudes  is  to  the  second  as 
the  second  is  to  the  third,  the  second  is  called  a  mean 
proportional  between  the  two  others. 

It  does  not  belong  to  geometry  to  develop  the  prin- 
ciples of  proportion ;  but,  inasmuch  as  they  are  ap- 
plied in  it,  some  of  the  more  important  of  them  are 
stated  here,  in  explanation  of  the  terms  and  phrases 
by  which  they  will  be  quoted.  None  will  be  referred 
to  that  are  not  found  in  common  arithmetics. 

3.  Multiplying  extremes  and  means,  is  the  phrase 
used  in  citing  the  principle  that  the  product  of  the 
extremes  of  a  proportion  is  equal  to  the  product  of 
the  means.     Thus, 

If  A  :  B  :  :  C  :  D,  then,  AxD=BxC. 

4.  Resolving  into  a  projyortion,  refers  to  the  prin- 
ciple that  if  the  product  of  two  magnitudes  is  equal 
to  the  product  of  two  others,  either  couple  may  be 
made  the  extremes  of  a  proportion  and  the  other  the 
means.     Thus, 

If  AxD=BxC,  then  A  :  B  :  :  C  :  D. 

5.  Multiplying  an  extreme  and  a  mean  equally, 
refers  to  the  principle  that  we  may  multiply  an 
extreme  and  a  mean  by  the  same  quantity  without 
destroying  the  proportion.     Thus, 

If  A  :  B  :  :  C  :  D,  then  A  :  2B  :  :  C  :  2D. 

6.  Dividing  an  extreme  and  a  mean  equally,  refers 
to  the  principle  that  we  may  divide  an  extreme  and  a 


BOOK    II.  57 

mean   by  the   same   quantity  without   destroying    the 
proportion.     Thus, 

If  A  :  B  :  :  C  :  D,  then  A  :  B  :  :  —  :  — . 

2      2 

7.  By  composition,  implies  that  if  any  number  of 
magnitudes  are  proportionals,  the  sum  of  all  the 
antecedents  is  to  the  sum  of  all  the  consequents  as 
any  one  antecedent  is  to  its  consequent.     Thus, 

If  A  :  B  :  :  C  :  D  :  :  E  :  F, 
Then   A-fC-f-E  :  B-f-D-f F  :  :  A  :  B. 

8.  By  alternation,  implies  that  the  order  of  the 
means  in  a  proportion  may  be  inverted.     Thus, 

If  A  :  B  :  :  C  :  D,  then  A  :  C  :  :  B  :  D. 

9.  By  equality  of  ratios,  implies  that  if  two  ratios 
are  respectively  equal  to  a  third,  they  are  equal  to 
each  other.      Thus, 

If   A  :  B  :  :  C  :  D    and    A  :  B  :  :  E  :  F, 
Then    C  :  D  :  :  E  :  F. 

10.  3Iultiplying   corresponding  terms,  is   the    phrase' 
used  in  applying  the  principle  that  the  product  of  the 
first  terms  of  two  proportions  is  to  the  product  of  the 
second   terms,  as   the    product    of  the  third  is  to  the 
product  of  the  fourth.     Thus, 

If  A  :  B  :  :  C  :  D,  and  E  :  F  :  :  G  :  H, 
Then  AxE  :  BxF  :  :  CxG  :  DxH. 

11.  Squaring  all  the  terms,  is  a  reference  to  tho 
principle  that  if  any  number  of  magnitudes  are  pro- 
portionals their  squares  are  also  proportionals.     Thus, 

If  A  :  B  :  :  C  :  D,  then  A2  :  B2  :  :  C2  :  D2. 


58 


GEO'METEY, 


THEOREM    I. 

If  two  triangles  have  the  same  altitude,  their  areas 
are  to  each  other  as  their  bases. 

A  T) 

Let  the  triangles 
ABC,  DEF,  have  the 
same  altitude ;  that 
is,  let  the  perpendic- 
ulars AG,  DH,  be 
equal.  It  is  to  be  proved  that  the  area  ABC  is  to  the 
area  DEF,  as  the  base  BC  is  to  the  base  EF. 

The  area  ABC  is  equal  to  -JAGxBC  (Cor.  2,  Theo. 
XVII,  Book  I) ;  and  the  area  DEF  is  equal  to 
JDHXEF.     Therefore, 

ABC  :  DEF  :  :  1-AGxBC  :  JDHxEF. 
But,   from   the    hypothesis,    JAG    is    equal    to    JDH. 
Hence,  dividing  a  mean  and  an  extreme  equally  (Def. 
6,  Sec.  X),  we  have 

ABC  :  DEF  :  :  BC  :  EF. 

Therefore,  if  two  triangles,  etc. 


THEOEEM    II. 

If  a  straight  line  be  drawn  parallel  to  the  base  of  a 
triangle  it  ivill  cut  the  other  sides  proportionally. 

In  the  triangle  ABC,  let  DE  be 
drawn  parallel  to  the  base  BC.  It 
is  to  be  proved  that  it  cuts  AB  and 
AC  proportionally. 

Join  BE  and  CD.  Now,  since  the 
triangles  ADE,  BDE,  have  the  same       B  c 

altitude,  namely,   the   perpendicular    distance    from   E 


BOOK    II.  59 

to  the  line  of  their  bases  AB  (Def.  4,  Sec.  V,  B.  I), 
their  areas  are  to  each  other  as  their  bases  (Theo. 
I) ;  that  is, 

ADE  :  DBE  :  :  AD  :  DB. 
Also,  because  the  triangles  ADE,  CDE,  have  the  same 

altitude, 

ADE  :  CDE  :  :  AE  :  EC. 

But  the  first  ratio  is  the  same  in  both  proportions ; 
for  ADE  is  common,  and  the  triangles  DBE,  CDE, 
having  the  same  base,  DE,  and  the  same  altitude, 
namely,  the  perpendicular  distance  between  the  par- 
allels DE  and  BC,  are  equivalent  (Theo.  I). 

Therefore,   by    equality    of  ratios  (Def.  9,  Sec.  X), 

we  have 

AD  :  DB  :  :  AE  :  EC; 

Or,  by  alternation  (Def.  8,  Sec.  X), 

AD  :  AE  :  :  DB  :  EC. 

Hence,  if  a  straight  line,  etc. 

Cor.  1.  By  composition  (Def.  7,  Sec.  X),  we  get 
from  the  last  proportion 

AD+DB  :  AE+EC  :  :  AD  :  AE, 
that  is,  AB  :  AC  :  :  AD  :  AE. 

Cor.  2.  If  DE  be  not  parallel  to  BC,  but  takes 
some  other  direction,  as  DF,  it  is  evident  that  it  will 
not  cut  the  other  sides  proportionally. 

THEOKEM    III. 
Two   triangles  ivliich   are   mutually   equiangular  are 
also  similar. 


60 


GEOMETRY. 


F 


Let  ABC,  BDE,  be  two  tri- 
angles having  the  angles  at  A, 
B,  and  C,  respectively  equal  to 
the  angles  at  B,  D,  and  E.  It 
is  to  be  proved  that  these  tri- 
angles are  similar. 

Let  them  be  so  placed  that 
their  homologous  sides  AB,  BD  (Def.  1,  Sec.  X),  shall 
form  one  straight  line.  Produce  AC  and  DE  till  they 
meet.  Then,  because  the  angle  DBE  is  by  hypothesis 
equal  to  the  corresponding  inner  angle  BAC,  the  lines 
BE  and  AC  are  parallel  (Cor.  1,  Theo.  Ill,  B.  I) ; 
and  because  the  angle  ABC  is  equal  to  the  corres- 
ponding inner  angle  BDE,  the  lines  BC  and  DE  are 
parallel ;  hence,  CBEF  is  a  parallelogram.  Now,  in 
the  triangle  ADF  we  have  (Theo.  II) 

DB  :  BA  :  :  DE  :  EF. 

But  EF  is  equal  to  BC  (Theo.  XIII,  B.  I).     Hence, 
DB  :  BA  :  :  DE  :  BC. 

By  a  like  construction  it  may  be  proved  in  the  same 
way  that  the  other  homologous  sides  are  proportional. 
The  two  triangles  are  consequently  similar  (Def.  1, 
Sec.  X). 

Therefore,  two  triangles,  etc. 

Cor.  If  two  triangles  have  two  angles  of  the  one 
respectively  equal  to  two  angles  of  the  other,  they  are 
similar ;  for  in  that  case  the  third  angles  are  neces- 
sarily equal  (Theo.  V,  B.  I). 


THEOREM   IV. 

Two  triangles  having  the  sides  of  the  one  successively 
proportional  to  the  sides  of  the  other,  are  similar. 


BOOK     II. 


Gl 


Let  ABC,  DEF,  be 
two  triangles  having 
AB  :  DE  :  :  BC  :  EF 
:  :  CA  :  FD.  It  may 
be  shown  that  they 
are  similar. 

On  DE  describe  the 
triangle  DGE,  having  the  angles  EDG,  DEG,  respect- 
ively equal  to  the  angles  A  and  B.  Then  will  the 
triangles  ABC  and  DEG  be  similar  (Cor.,  Theo.  Ill) ; 
and  we  shall  have  (Def.  1,  Sec.  X) 

AB  :  DE  :  :  BC  :  EG. 

But  by  hypothesis  we  have 

AB  :  DE  :  :  BC  :  EF. 

Hence,  EG  is  equal  to  EF.  In  the  same  manner  it 
may  be  shown  that  DG  is  equal  to  DF.  Therefore, 
the  two  triangles  DGE,  DFE,  are  mutually  equilateral 
and  consequently  equal  (Cor.,  Theo.  XII,  B.  I).  But 
ABC  is  by  construction  similar  to  DGE :  hence,  it  is 
also  similar  to  DFE. 

Therefore,  two  triangles,  etc. 


THEOEEM    V. 

Two  triangles  having  an  angle  of  the  one  equal  to  an 
angle  of  the  other,  and  the  sides  about  those  angles  pro- 
portional,  are  similar. 

Let  the  two  triangles 
ABC,  DEF,  have  the 
angle  A  equal  to  the 
angle  D,  and  the  sides 
AB,  AC,    proportional   to 


62 


GEOMETRY. 


the    sides    DE,    DF.     Then   will    these    triangles    be 
similar. 

Take  AG  equal  to  DE,  and  AH  to  DF ;  also,  join 
GH.  Then  the  triangles  AGH,  DEF,  having  two 
sides  and  the  included  angle  of  the  one  equal  to  two 
sides  and  the  included  angle  of  the  other,  are  equal 
throughout  (Theo.  VII,  B.  I).  Now,  by  hypothesis, 
AB  :  DE  :  :  AC  :  DF. 

Therefore,  AB  :  AG  :  :  AC  :  AH. 
Hence,  it  follows  that  GH  is  parallel  to  BC  (Cor.  2, 
Theo.  II).  Consequently  the  angle  AGH  is  equal  to 
the  angle  ABC  (Theo.  Ill,  B.  I),  and  AHG  to  ACB. 
Consequently,  also  (Cor.,  Theo.  Ill),  the  triangle 
AGH  or  its  equal  DEF  is  similar  to  ABC. 

Therefore,  two  triangles,  etc. 

Cor.  If  a  triangle,  as  AGH,  be  cut  off  from  an- 
other triangle,  ABC,  by  a  straight  line  parallel  to  the 
base,  the  two  triangles  will  be  similar. 


THEOREM  VI. 

Two  triangles  having  their  sides  respectively  parallel 
or  perpendicular  to  each  other,  are  similar. 

First,  let  the  tri- 
angles ABC,  abc,  have 
their  sides  respect- 
ively parallel.  Then, 
because  the  sides  con- 
taining the  angle  A 
are  parallel  to  the 
sides  containing  the 
angle    a,  these    angles   are    equal  (Cor.  2,  Theo.  Ill, 


BOOK    II. 


63 


B.  I) :  and  for  the  same  reason  the  angles  B  and 
b  are  equal,  also  C  and  c.  Hence,  the  two  triangles 
are  mutually  equiangular,  and  consequently  similar 
(Theo.  III).     . 

Secondly,  let  the  triangles  ABC,  DEF,  have  their 
sides  respectively  perpendicular  to  each  other.  Pro- 
duce the  sides  of  DEF  to  the  points  G,  H,  I.  Now, 
the  sum  of  all  the  angles  of  the  quadrilateral  AIDG 
is  equal  to  four  right  angles  (Theo.  XX,  B.  I).  But 
AID  and  AGD  are  right  angles  by  hypothesis. 
Hence,  IDG  and  IAG  must  be  together  equal  to  two 
right  angles.  But  IDG  and  EDF  are  also  together 
equal  to  two  right  angles  (Theo.  I,  B.  I).  Therefore, 
subtracting  equals  from  equals,  we  have  the  angle 
IAG  equal  to  the  angle  EDF.  In  the  same  manner 
it  may  be  proved  that  the  angle  ACB  is  equal  to  the 
angle  DFE,  and  CBA  to  FED.  Hence,  the  two  tri- 
angles are  mutually  equiangular,  and  consequently 
similar. 

Therefore,  two  triangles,  etc. 


THEOEEM    VII. 

A  perpendicular  let  fall  from  the  right  angle  iipon 
the  hypotenuse  of  a  rigid-angled  triangle  divides  it  into 
two  triangles  similar  to  the  whole,  and  to  each  other. 

Let  ABC  be    a  triangle  right- 

angled  at  B,  and  let  BD  be  per- 
pendicular to  the  hypotenuse  vVC. 
It  is  to  be  proved  that  ABD  and 
BCD  are  similar  to   ABC  and  to 

each  other. 

The    triangles   ABD,   ABC,   have   a  right  angle   in 


64 


GEOMETRY. 


^u> 


each,  and  the  angle  A  common ;  hence,  they  are 
gar  (Cor.,  Tlieo.  III).  The  triangles  BCD,  ABC, 
e  also  a  right  angle  in  each,  and  the  angle  C 
common ;  and  hence  they  are  similar.  Therefore, 
also,  the  triangles  ABD,  BCD,  being  both  similar 
to  the   same  triangle,  arcj^Ailar  to  each  other. 

Hence,  a  perpendffl  mu 

Cor.      By  similaritjjB  JjiLangh's  ADB,  BDC,  we 

have  AD  :  DB  :  :  DB  :^^;  that  is,  the  perpendicular 
from  the  right  angle  is  a  mean  proportional  between  the 
parts  of  the  hypotenuse  (Def.  2,  Sec.  X). 


THEOREM   VIII. 


The  areas   of  similar  triangles   are   to   each   other  as 
the  squares  described  on  their  homologous  sides. 


D 


E 


/L 


H 


T 


Let  ABC,  DEF,  be 

two   similar    triangles 
of   which    the    angles 
A  and  B  are  respect- 
ively equal  to  the  an-    R 
gles    D    and    E.      It 

may  be   shown  that  their  areas  are   to  each  other  as 
AB2  is  to  DE2  (Def.  1,  Sec.  X). 

Draw  the  perpendiculars  AG'  and  DH.  Then,  the 
triangles  ABG,  DEH,  having  a  right  angle  in  each, 
and  the  angles  B  and  E  equal,  are  similar  (Cor., 
Theo.  Ill),  and  Ave  have 

GA  :  HD  :  :  AB  :  DE. 
But  by  similarity  of  ABC  and  DEF  we  have 
BC  :  EF  :  :  AB  :  DE. 


BOOK    II.  65 

Multiplying  corresponding  terms  (Def.  10,  Sec.  X) 
and  dividing  an  extreme  and  a  mean  equally  ilK'^^i 
Sec.  X),  ^^ 

BCXGA  :  EFXHD  DE2 

2  2 

But  the  first  two  terms  ofl|  ■&  proportion  represent 
the    areas   of   the    triaii^|j  |£  and  DEF  (Cor.  2, 

Theo.  XVII,  B.  I). 

Therefore,  the  areas, 


THEOREM    IX. 

If  between  the  tivo  parallel  sides  of  a  trapezoid  a 
third  parallel  be  drawn,  it  will  cut  the  other  sides  pro- 
portionally. 

In  the  trapezoid  ABCD,  r                      —\ 

let  EG  be  a  third  parallel  /               „-'"'        \ 

to  AD  and  BC.     Then  will  J    ^-'  '' V 

BE  be  to  EA  as  CG  is  to  L^- A, 

GD. 

Draw  the  diagonal  BD.  Now,  because.  EF  is  par- 
allel to  the  base  AD  of  the  triangle  BAD,  we  have 

(Theo.  II) 

BE  :  EA  :  :  BF  :  FD ; 

and,  because    FG   is   parallel    to  the  base  BC   of  the 

triangle  DBC, 

CG  :  GD  :  :  BF  :  FD. 

Therefore,  by  equality  of  ratios  (Def.  9,  Sec.  X), 

BE  :  EA  :  :  CG  :  GD. 

Hence,  if  between,  etc. 

Cor.  1.     If  the  third  parallel  bisects  the  oblique  sides, 
it  is  equal  to   one-half  the   sum  of  the  parallel  sides. 
Evans    Geometry. — 6 


66 


GEOMETKY. 


For  by  similarity  of  the  triangles  BEF,  BAD  (Cor. 
Theo.  V),  if  BE  be  one-half  of  BA,  EF  will  be  one- 
half  of  AD ;  and  in  the  same  manner  it  may  be 
shown  that  FG  will  be  one-half  of  BC  :  consequently, 
the  whole  line  EG  will  be  one-half  the  sum  of  AD 
and  BC. 

Cor.  2.  The  area  of  the  trapezoid  (Theo.  XVIII, 
B.  I)  is  equal  to  its  altitude  multiplied  by  EG,  which 
is  its  mean  breadth. 


THEOREM   X. 

The  perimeters  of  similar  polygons  are  to  each  other 
as  their  homologous  sides.  >» 

Let  ABCDE, 
abcde,  be  two  simi- 
lar polygons,  hav- 
ing the  angles  A, 
B,  C,  etc.,  respect- 
ively equal  to  the 
angles   a,  b,  c,  etc. 

Then  will   their  perimeters   be   to    each   other  as  any 
two  homologous  sides  AB  and  ab. 

By  similarity  of  the  two  polygons  (Def.  1,  Sec.  X) 
we  have 

AB  :  ab  :  :  BC  :  be  :  :  CD  :  cd  :  :  DE  :  de  :  :  EA  :  ea. 
And  therefore,  by  composition  (Def.  7,  Sec.  X), 
AB+BC-f-CD  +  DE  +  EA  :  ab+bc+cd+de+ea  :  : 

AB  :  ab. 
But  the  first  two  terms  of  the  last  proportion  repre- 
sent the  perimeters  of  the  two  polygons  (Def.  3,  Sec. 
VII,  B.  I.) 

Therefore,  the  perimeters,  etc. 


BOOK    II. 


67 


THEOREM   XL 

Tlie  areas  of  two  regular  polygons  of  the  same  num- 
ber of  sides  are  to  each  other  as  the  squares  of  their 
sides. 

Let  ABODE, 
abode,  be  two 
regular  polygons  -g 
of  the  same  num- 
ber of  sides,  for 
example,  five. 
The  area  of  the 
former  is  to  the  area  of   the  latter  as  AB2  is  to  air. 

Since  the  two  polygons  have  the  same  number  of 
sides  and  angles,  it  is  evident  (Theo.  XX,  B.  I)  that 
the  sum  of  all  the  angles  of  ABODE  is  equal  to  the 
sum  of  all  the  angles  of  abode,  and  consequently  that 
each  angle  of  ABODE  is  equal  to  each  angle  of 
abode  (Del  2,  Sec.  VII). 

Bisect  the  angles  A  and  B  by  AF  and  BF,  and  the 

angles  a  and  b  by  af  and  bf.     Now,  since  the  triangles 

ABF,  abf,  have  two    angles   of   the  one  equal  to  two 

angles  of  the  other,  they  are  similar  (Cor.,  Theo.  Ill)  ; 

hence 

ABF  :  abf  :  :  AB2  :  ab°  (Theo.  VIII). 

Multiplying  an  extreme  and  a  mean  equally, 

5  ABF  :  babf  :  :  AB2  :  ab2  (Def.  5,  Sec.  X). 
But   five    times   ABF   is    the    area  of  ABODE  (Theo. 
XXI,  B.  I),  and  five  times  abf  is  the  area  of  abode. 

Therefore,  the  areas,  etc. 


68  G-EOMETRY. 

EXEECISES. 

1.  Prove  that  two  parallelograms  of  the  same  alti- 
tude are  to  each  other  as  their  bases. 

2.  Prove  that  if  two  isosceles  triangles  have  their 
vertical  angles  equal  they  are  similar. 

3.  If  the  area  of  a  regular  pentagon  whose  side  is 
1  be  1.72,  what  is  the  area  of  a  regular  pentagon 
whose  side  is  5? 

4.  In  two  similar  polygons,  if  a  side  of  one  be  7, 
and  the  homologous  side  of  the  other  13,  and  if  the 
perimeter  of  the  former  be  39,  Avhat  is  the  perimeter 
of  the  latter? 

SEC.  XI.— OF   CIRCLES. 
DEFINITIONS. 

1.  Two  ARCS  of  circles  are  called  similar  when  they 
are  equal  parts  of  the  circumferences  to  which  they 
belong. 

2.  Two  sectors  are  called  similar  when  their  arcs 
are  similar. 

THEOREM   XII. 

If  two  chords  intersect  in  a  circle,  the  product  of  the 
parts  of  the  one  is  equal  to  the  product  of  the  p>arts  of 
the  other. 

In  the    circle  ABCD  let  the  D/^  "^Xn 

chords  AC,  BD,  intersect  each  //    ^\      ^s\ 

other  in  E.     It  is  to  be  proved  [  /      ^-^e"\       •  j 

that  AExEC=BExED.  \L^  ^nI 

Join  AD  and  BC.     Now,  in  \  / 

the    triangles  ADE,  BCE,    the  ^ ^ 

vertical  angles   AED  and   BEC  are  equal  (Theo.  II, 


BOOK    II. 


69 


B.  I) ;  also,  the  angles  A  and  B  are  equal,  being  both 
measured  by  half  the  same  arc,  DC  (Theo.  XXVI,  B. 
I) ;  hence,  the  third  angles  are  equal,  and  the  two 
triangles  are  similar  (Theo.  Ill),  and  we  have 

AE  :  BE  :  :  ED  :  EC. 
Then,  multiplying  extremes  and  means,  we  get 
AExEC=BExED. 

Therefore,  if  two  chords,  etc. 


THEOREM  XIII. 

If  from  a  point  without  a  circle  a  tangent  and  a 
secant  he  drawn,  the  tangent  ivill  be  a  mean  'propor- 
tional between  the  secant  and  its  external  part. 

Let  AB  be  a  tangent  and  AC 
a  secant  to  a  circle.     Then  will 
AC  :  AB  :  :  AB  :  AD. 

Join  BC  and  BD.  Now,  the 
angle  DBA,  contained  by  a  tan- 
gent and  chord,  is  measured  by 
half  the  arc  DB  (Theo.  XXVII, 
B.  I) ;  and  the  angle  C,  being  an 
inscribed  angle,  is  measured  by  half  the  same  arc 
(Theo.  XXVI,  B.  I) ;  therefore,  these  two  angles  are 
equal.  But  the  angle  A  is  common  to  the  two  tri- 
angles BAD  and  BAC.  Hence,  these  triangles  have 
two  angles  of  the  one  equal  to  two  angles  of  the 
other,  and  are  consequently  similar  (Cor.,  Theo.  III). 

Therefore,  AC  :  AB  :  :  AB  :  AD. 
That  is,  if  from  a  point,  etc. 


70  GEOMETRY. 

THEOREM    XIV. 

Similar  arcs  in  different  circles  subtend  equal  angles 
at  the  centers. 

Let  AB  and  ah  be 
similar  arcs  -in  two 
circles  whose  centers 
are  C  and  c.  It  may- 
be shown  that  the  an- 
gles at  C  and  c  are  equal. 

Whatever  portion  AB  is  of  the  circumference  to 
which  it  belongs,  the  angle  C  is  the  same  portion  of 
four  right  angles  (Schol.,  Theo.  XXIV,  B.  I);  and 
whatever  portion  ah  is  of  the  other  circumference,  the 
angle  c  is  the  same  portion  of  four  right  angles.  But 
AB  is  the  same  portion  of  the  first  circumference  that 
ab  is  of  the  second  (Def.  1,  Sec.  XI).  Hence,  the 
angles  C  and  c  are  equal  portions  of  four  right 
angles ;  hence,  also,  they  are  equal  to  each  other. 

Therefore,  similar  arcs,  etc. 

Cor.  The  sectors  ABC,  abc,  are  similar  (Def.  2, 
Sec.  XI).  Hence,  the  sides  of  similar  sectors  con- 
tain equal  angles.  Hence,  also,  similar  sectors  are 
contained  an  equal  number  of  times  in  the  circles  to 
which  they  belong. 

Schol.  Since  a  degree  is  ^^  of  the  circumference 
in  which  it  is  taken  (Schol.,  Theo.  XXIV,  B.  I),  it 
follows  that  degrees  in  unequal  circles  are  similar  but 
not  equal  arcs. 


BOOK     II. 


71 


THEOEEM   XV. 

The  circumferences  of  two   circles  are  to  each  other 
as  their  diameters. 

J 


C    / 


Let  ACE 
and  ace  be  two 
circles  having 
two  similar 
regular  poly- 
gons inscribed, 
one     in     each. 

Draw  the  radii  AG,  BG ;  ag,  bg.  Now,  the  angles  G  and 
g,  being  subtended  by  similar  arcs,  AB  and  ab,  are 
equal  (Theo.  XIV) ;  and  the  sides  containing  them  are 
by  construction  proportional ;  therefore,  the  two  tri- 
angles AGB,  agb,  are  similar  (Theo.  V),  and  we  have 

AB  :  ab  :  :  AG  :  ag  :  :  2  AG  :  2  ag. 

But  the  perimeter  ABCDEF  is  to  the  perimeter 
abcdef,  as  AB  is  to  ab  (Theo.  X),  and  consequently 
as  2  AG  is  to  2ag.  Now,  if  the  number  of  sides  of 
the  similar  polygons  be  indefinitely  increased,  the 
perimeters  will  ultimately  coincide  with  the  circum- 
ferences.    Therefore, 

Circumference  ACE  :  circumference  ace  :  :  2  AG  :  2  ag. 
But  2  AG  and  2  ag  represent  the  diameters  of  the  two 
circles  (Def.  2,  Sec.  8,  B.  I). 

Therefore,  the  circumferences,  etc. 

Cor.  1.     Representing  the  diameter  of  any  circle  by 

D,  and  its  circumference  by  C,  and  remembering  that 

the  circumference   of  a  circle  whose  diameter  is  1   is 

3.14159  (Schol.,  Theo.  XXX,  B.  I),  we   have,  by  the 

above  theorem, 

1   :  D  :  :   3.14159  :  C. 


72 


GEOMETRY. 


Multiplying  extremes  and  means,  we  get  C  =  3.14159 
XD.     That  is, 

Tlic  circumference  of  any  circle   is  equal   to    its  di- 
ameter multiplied  by  3.14159. 

Cor.  2.     The  diameter  is  equal  to  the  circumference 
divided  by  3.14159. 


THEOREM    XYI. 

TJie  areas   of  two   circles   are   to   each  other   as  the 
squares  of  their  diameters. 

The  same  con- 
struction being 
used  as  in  the 
last  theorem,  it 
may  be  shown 
as  before,  that 

AB  :  ab  :  :  2  AG  :  2  ag. 

Hence,  by  squaring  all   the   terms  (Def.  11,  Sec.  X), 

AB-  :  ab2  :  :  (2  AG)2  :  (2ag)2. 
But  the  area  of  the  polygon  ABCDEF  is  to  the  area 
of  the  polygon  abedef,  as  AB2  is  to  ab2  (Theo.  XI), 
and,  consequently,  as  (2  AG)2  is  to  (2  ag)2.  Now,  if 
the  number  of  sides  of  the  two  polygons  be  in- 
definitely increased,  their  areas  will  ultimately  coincide 
with  the  areas  of  the  circles.  Hence,  the  area  of  the 
circle  ACE  is  to  the  area  of  the  circle  ace  as  (2  AG)2 
is  to  (2ag)2. 

That  is,  the  areas  of  two  circles,  etc. 

Cor.     Representing   the  diameter  of   any  circle    by 
D  and  its  area  by  A,  and  remembering  that  the  area 


BOOK    II.  73 

of  a  circle  whose  diameter  is  1  is  .7854  (Cor.,  Theo. 
XXX,  B.  I),  we  have 

V  :  D2  :  :  .7854  :  A. 
But    I2   is    1.     Therefore,    multiplying    extremes    and 
means,  we  get  A  =  .7854xD'2.     That  is. 

The  area  of  any  circle  is  equal  to  the  square  of  its 
diameter  multiplied  by  .7854. 

EXEKCISES. 

1.  If  a  secant  to  a  circle  be  12  feet,  and  its  external 
part  8  feet,  what  will  be  the  length  of  a  tangent  to 
the  circle,  drawn  from  the  same  point  ? 

2.  If  the  diameter  of  a  circle  be  24,  Avhat  will  be 
the  length  of  its  circumference? 

3.  If  the  circumference  of  a  circle  be  8924,  what  is 
the  length  of  its  diameter  ? 

4.  The  diameter  of  a  circle  is  16:  what  its  area? 

5.  Prove  that  the  areas  of  similar  sectors  are  to 
each  other  as  the  squares  of  their  radii. 

6.  Prove  that  the  area  of  a  circle  equals  the  square 
of  its  radius  multiplied  by  3.14159. 

SEC.  XII.— PROBLEMS   IN   CONSTRUCTION. 

PROBLEM    I. 
To  divide  a  straight   line  into  any  number  of  equal 
parts. 

Solution.  Let  AB  be 
the  given  straight  line, 
which    it    is    required    to 

divide  into  a  certain  num-     A-^ j^ 

ber  of  equal  parts,  for  example,  three 
Evans'  Geometry. — 7 


O  B 


74 


GEOMETRY. 


From  A  draw  any  straight  line  AD ;  and  lay  off 
DE  and  EC  in  the  same  direction,  each  equal  to  AD. 
Join  CB,  and  through  D  draw  DF  parallel  to  CB. 
Now,  AD  :  AC  :  :  AF  :  AB  (Cor.  1,  Theo.  II).  But 
AD  is  one-third  of  AC,  by  construction ;  therefore, 
AF  is  one-third  of  AB.  Now,  lay  off  FG  equal  to 
AF,  and  it  is  evident  that  AB  will  be  divided  into 
three  equal  parts. 

PROBLEM  II. 

To  find  a  fourth  proportional  to  three  given  straight 
lines. 

Solution.  Draw  two  straight 
lines  containing  any  angle  A. 
Make  AB,  BC,  and  AD  re- 
spectively equal  to  the  three 
given  lines,  and  join  BD. 
Through  C  draw  CE  parallel  to  BD,  and  produce  AD 
to  meet  it  in  E.  Now,  AB  :  BC  :  :  AD  :  DE  (Theo. 
II).  Hence,  DE  is  the  fourth  proportional  required 
(Def.  2,  Sec.  X). 

PROBLEM   III. 

To  find  a  mean  proportional  between  two  given 
straight  lines. 

Solution.  Lay  off,  in  one 
straight  line,  AB  and  BC, 
respectively  equal  to  the 
two  given  lines.  Find  E, 
the  middle  point  of  AC. 
Then,  from  E  as  a  center,  with  EA  or  EC  as  radius, 
describe  a  semicircle.  At  B  erect  BD  perpendicular 
to  AC,  and  join  AD  and  DC.     Now,  since  the  angle 


A1-' 


BOOK    II.  75 


ADC  is  inscribed  in  a  semicircle,  it  is  a  right  angle 
(Cor.  2,  Theo.  XXVI,  B.  I).  Hence,  BD  is  a  mean 
proportional  between  AB  and  BC  (Cor.,  Theo.  YII). 

PROBLEM   IV. 

To  describe  a  square  that  shall  be  equivalent  to  a 
given  parallelogram. 

Solution.     Let    ABCD    be    the 

given  parallelogram.     Between  its  /• 7 

base  AB  and  its  altitude  ED,  find 
a  mean  proportional,  which  denote 
by  M.     Then,  since  A    E  E 

AB  :  M  :  :  M  :  ED,  we  have  ABxED=M2. 

But  ABxED  represents  the  area  of  the  parallelogram 
(Cor.  1,  Theo.  XVII,  B.  I).  Therefore,  the  square 
described  on  M  is  the  square  required. 

Cor.  It  is  evident  that  if  we  find  a  mean  propor- 
tional between  the  base  of  a  triangle  and  half  its 
altitude  (Cor.  2,  Theo.  XVII,  B.  I),  we  shall  have  the 
side  of  a  square  equivalent  to  the  triangle. 

EXERCISES. 

1.  To  construct  a  triangle  that  shall  be  similar  to  a 
given  triangle. 

2.  On  a  given  straight  line,  to  describe  a  rectangle 
that  shall  be  equivalent  to  a  given  square. 


BOOK    III, 


SOLID  GEOMETRY 


SEC.  XIII.— PLANES  AND  THEIR  INCLINATIONS. 


DEFINITIONS. 

1.  A  straight  line  is  said  to  be  perpendicular  to  a 
plane  when  it  is  perpendicular  to  every  straight  line 
it  can  meet  in  that  plane. 

2.  The  line  in  -which  two  planes  cut  one  another  is 
called  their  intersection. 

Schol.  The  intersection  of  two 
planes  is  a  straight  line.  For, 
if  A  and  D  are  any  two  points 
in  the  intersection  of  the  planes 
FC  and  EB,  the  straight  line 
joining  these  points  must  be  in 
both  planes  (Def.  8,  Sec.  I) ;  it  is, 
therefore,  their  intersection. 

3.  The  angle  of  inclination  of  two  planes  is  that 
contained  by  any  two  straight  lines  perpendicular  to 
their  intersection,  one  in  each  plane. 

If  the   line   in   each    plane   is    perpendicular  to  the 
other  plane,  the  two  planes  are  said  to  be  perpendicu- 
lar to    each   other,  the    angle    of   inclination    being    a 
right  angle. 
(76) 


BOOK   III. 


77 


Thus,  if  AB  and  CB'  are  both  perpendicular  to  the 
intersection  EH,  then  ABC  is 
the  angle  of  inclination  of  the 
planes  DII  and  HF.  If  AB  is 
also  perpendicular  to  the  plane 
HF,  and  CB  to  DII.  the  two  D< 
planes  are  perpendicular  to  each 
other. 

4.  Two  planes  are  said  to  be 
parallel  with  one  another  when  their  intersections  by 
any  third  plane  are  parallel. 

Cor.  Two  parallel  planes  can  never  meet ;  for,  if 
they  were  to  meet  in  any  point,  their  intersections  by 
a  third  plane  passing  through  that  point  would  also 
meet;   which  is  contrary  to  the  definition. 

5.  The  divergence  of  three 
or  more  planes  from  a  single 
point  constitutes  a  solid  an- 
gle. Thus,  A  is  a  solid  an- 
gle contained  by  the  three  A 
planes  BAC,  CAD,  DAB. 


THEOKEM    I. 

Two   straight  lines  perpendicular  to  the  same  plane 
are  parallel  to  each  other. 

Let  AB  and  CD  be  two 
perpendiculars  to  the  plane 
EF.  It  is  to  be  shown  that 
they  are  parallel. 

Join  BD.  Now,  AB  and 
CD,  being  perpendicular  to 
the    plane,    are    also    perpendicular   to    the    line    BD 


78 


(tEOMETR  y. 


"which  they  meet  in  it  (Def.  1).  Also,  a  plane  per- 
pendicular to  EF  and  passing  through  BD  will  contain 
both  AB  and  CD  (Def.  3).  But  straight  lines  in  the 
same  plane  and  perpendicular  to  the  same  straight  line 
are  parallel  (Cor.  2,  Theo.  IV,  B.  I).  Therefore,  AB 
is  parallel  to  CD. 

That  is,  two  straight  lines,  etc. 


THEOREM   II. 

-  Parallel  lines  intercepted  between  parallel  planes  are 
equal. 

Let  FH  and  EG  be  any 
two  parallel  lines  intercepted 
between  two  parallel  planes, 
AB  and  CD.  It  is  to  be 
shown  that  FH  is  equal  to 
EG. 

Through  the  parallels  FH 
and  EG  let  the  plane  FG  pass.  Its  intersections  with 
the  parallel  planes  AB  and  CD  will  be  parallel  lines 
(Def.  4) ;  that  is,  FE  is  parallel  to  HG.  Hence, 
FEGH  is  a  parallelogram,  and  FH  is  equal  to  EG 
(Theo.  XIII,  B.  I). 

Therefore,  parallel  lines,  etc. 

Cor.  1.     Any  straight  line,  as  FH,  perpendicular  to 

one  of  two  parallel  planes,  as  CD,  will  be  perpendicular 
to  the  other.  For  FHG  being  a  right  angle  (Def.  1), 
HFE  will  also  be  a  right  angle  (Cor.  1,  Theo.  IV, 
B.  I),  whatever  may  be  the  position  of  the  parallels 
FE  and  HG  in  their  planes. 

Cor.  2.     The    perpendicular   distance    between    two 
parallel  planes  is  everywhere  the  same. 


BOOK    III. 


79 


SEC.  XIV.— OF  PARALLELOPIPEDS. 

1.  A  PAiiALLELOPiPED  is  a  solid  bounded  by  six 
plane  faces,  of  which  each  one  E 

is  parallel  to  its  opposite.  The 
intersections  of  the  faces  are 
called  edges.         r 

Cor.  Any  face  of  a  parallele- 
piped is  a  parallelogram.  For 
the  lines  GH  and  BC,  being  the  intersections  of  two 
parallel  planes,  FH  and  AC,  with  a  third  plane  GC,  are 
parallel  (Def.  4,  Sec.  XIII);  and  in  the  same  way  it 
may  be  shown  that  BG  and  CII  are  parallel ;  hence, 
BCHG  is  a  parallelogram  (Def.  2,  Sec.  VI,  B.  I). 

2.  A  right  parallelopiped  is  one  whose  edges  are 
perpendicular  to  the  faces.  Any  other  parallelopiped 
is  called  oblique. 

S.  A  cube  is  a  right  parallelopiped  whose  length, 
breadth,  and  hight  are  equal. 

4.  Two  parallelopipeds,  or  other  solids,  are  called 
equivalent  when  they  contain  the  same  amount  of  solid 
space. 


E 


D 


II 


THEOREM  III. 

Any  hvo  opposite  faces  of  a  parallelopiped  are  equal. 

Let  AH  be  a  parallelopiped. 
It  is  to  be  proved  that  any 
two  opposite  faces,  as  DEFA, 
CHGB,  are  equal.  Join  DF 
and  CG.  Now,  since  EFGH  is 
a  parallelogram  (Cor.  Def.  1),  EF  is  equal  to  HG 
(Theo.  XIII,  B.  I) ;  and  in  the  same  manner  it  may  be 
shown  that  DE  is  equal  to  CH.     But   FG  and   DC, 


\ 

\ 

F 

/ 

/ 

/ 

C 

/ 

\ 

\ 

c; 


J: 


80 


GEOMETRY. 


being  both  equal  and  parallel  to  EH  (Theo.  XIII, 
B.  I),  must  be  equal  and  parallel  to  each  other.  Hence, 
DCGF  is  a  parallelogram  (Theo.  XIV,  B.  I),  and  DF 
is  equal  CG.  Wherefore,  the  triangles  DEF,  CHG,  are 
mutually  equilateral,  and  consequently  equal.  But  the 
former  triangle  is  half  the  parallelogram  DEFA  (Cor. 
1,  Theo.  XIII,  B.  I),  and  the  latter  triangle  is  half  the 
parallelogram  CHGB  :  hence,  these  two  parallelograms 
are  equal. 

Therefore,  any  two  opposite  faces,  etc. 

Schol.  Any  face  of  a  parallelopiped  may  be  taken 
as  the  base.  We  may  designate  DABC  and  EFGH  as 
the  lower  and  upper  bases.  The  perpendicular  distance 
between  the  bases  is  called  the  hight  or  altitude. 

Cor.  1.  If  EB  be  a  right  parallelopiped,  the  edges 
AF  and  BG,  since,  by  definition,  they  are  both  perpen- 
dicular to  the  face  DB,  must  be  perpendicular  also  to 
the  straight  line  AB,  which  they  meet  in  that  plane 
(Def.  1,  Sec.  XIII).  Hence,  ABGF,  or  any  face  of  a 
rigid  parallelopiped,  is  a  rectangle. 

Cor.  2.     The  faces  of  a  cube  are  all  equal  squares. 


THEOREM   IV. 

The  solidity  of  a  rigid  'parallelopiped  is  equal  to  the 
area  of  its  base  multiplied  by  its  hight. 

Let  ABC  be  a  right  paral- 
lelopiped. It  is  to  be  proved 
that  its  solidity  is  equal  to  the 
area  of  its  base  AB,  multiplied 
by  its  hight  DC. 

Let  planes  pass  through  the 
solid  parallel  to  the  three  faces' 


'  /  / 

/  /  / 

/ 

/ 

/ 

/ 
/ 

/ 
/ 

D 

BOOK    III. 


81 


AB,  BC,  CA,  dividing  the  edges  DA,  DB,  DC,  into  parts 
of  equal  length.  It  is  evident  that  the  parallelopiped 
will  thus  be  divided  into  a  number  of  equal  cubes.  Let 
one  of  these  be  taken  as  the  unit  of  solidity.  Now, 
in  the  layer  next  to  the  base,  there  will  be  as  many  of 
these  solid  units  as  there  are  corresponding  units  of  area 
in  the  base  ;  and  there  will  be  as  many  equal  layers  as 
there  are  corresponding  linear  units  in  the  hight. 

Therefore,  the  solidity  of  a  right  parallelopiped,  etc. 

Schol.  If  the  edges  are  incommensurable,  so  that 
no  unit  can  be  found  into  which  they  can  all  be  divided 
without  remainder,  the  above  theorem  will  still  hold 
true ;  for  by  taking  the  unit  smaller  and  smaller,  the 
remainder  can  be  made  less  than  anv  assignable 
quantity. 

When  the  linear  unit  is  one  inch,  the  unit  of  solidity 
is  a  cubic  inch  ;  when  the  linear  unit  is  one  foot,  the 
unit  of  solidity  is  a  cubic  foot,  etc. 

Cor.  1.  The  solidity  of  a  right  parallelopiped  is 
equal  to  the  product  of  its  length,  breadth,  and  hight. 

Cor.  2.  The  solidity  of  a  cube  may  be  found  by 
raising  one  of  its  edges  to  the  third  power. 


THEOKEM   V. 

Parallelopipeds   on   the   same    base   and   between   the 
same  parallels  are  equivalent. 

J) E  M K 

Let  AGHD  and  AGLM 

be  two  parallelopipeds  on 
the    same    base    AG,  and 
between  the   same    paral- 
lels  BG,    CL;    AF,  DK.        b  o 
It   may   be   shown    that   they    are    equivalent. 


82  GEOMETRY. 

The  parallelograms  AC,  FH,  being  opposite  faces 
of  a  parallelopiped,  are  equal  (Theo.  III).  Also,  the 
parallelograms  DH,  ML,  being  both  equal  to  AG,  are 
equal  to  each  other ;  hence,  if  we  add  EN  to  both, 
the  parallelograms  DN  and  EL  must  be  equal.  It  is 
also  evident  that  the  side  CN  is  equal  to  the  side  IIL. 
But  CB  is  equal  to  HG,  and  the  angle  BCN  is  equal 
to  the  corresponding  outer  angle  GHL  ;  therefore, 
the  triangles  BCN,  GHL,  having  two  sides  and  the 
included  angle  of  the  one  equal  to  two  sides  and 
the  included  angle  of  the  other,  are  equal  (Theo. 
VII,  B.  I). 

If  now  the  solid  FGLE  be  applied  to  the  solid 
ABND,  so  that  the  face  EL  shall  coincide  with  its 
equal  DN,  then  the  faces  GHL  and  BCN,  being  equal 
and  in  the  same  plane,  will  coincide ;  and  in  the  same 
manner  it  may  be  shown  that  the  faces  FEK  and 
ADM  will  coincide ;  hence,  also,  the  face  FH  will 
coincide  with  the  face  AC,  and  FL  with  AN ;  and  the 
two  solids  are  consequently  equal.  If,  then,  from  the 
whole  solid  AL,  we  take  away  these  equal  portions  in 
turn,  the  remainder  AGHD  will  be  equivalent  to  the 
remainder  AGLM. 

Therefore,  parallelopipeds  on  the  same  base,  etc. 

Schol.     Any    two    par-      V- XM E 


allelopipeds    on    the    same      x\ \  \  \ 

base  and  between  the  same 

D' \E 

planes      are      equivalent.  \  V 

For,  if   their  upper  bases  c  H 

are   not  between  the   same    parallels,  but   have    some 

other  position,  as   DH  and  MS,  produce  two  sides  of 

each  till  they  intersect  as  in  the  figure  ;  then  UVWX 


BOOK    III. 


83 


can  be  the  upper  base  of  a  parallelopipecl  on  the  same 
lower  base  and  between  the  same  parallels  with  both 
the  other  parallelopipeds ;  hence,  by  the  above  demon- 
stration, both  the  others  will  be  equivalent  to  this  third 
parallelopiped ;  they  will,  therefore,  be  equivalent  to 
each  other. 

Cor.  The  solidity  of  any  parallelopiped  on  a  rect- 
angular base,  is  equal  to  the  area  of  its  base  multi- 
plied by  its  altitude  ;  for  it  is  equivalent  to  a  right 
parallelopiped  of  the  same  base  and  altitude. 


SEC.  XV.— THE  PRISM  AND  THE  CYLINDER. 
DEFINITIONS. 

1.  A  prism  is  a  solid  bounded  by  plane  faces,  of 
which  two  are  equal  and  parallel  polygons,  and  the 
Others  parallelograms.  It  includes  the  parallelopiped, 
as  one  species. 

The  equal  and  parallel  polygons 
are  called  the  bases.  The  other 
faces  together  form  the  convex  or 
lateral  surface.  The  edges  join- 
ing the  corresponding  angles  of  the 
two  bases  are  called  the  principal 
edges. 

A  prism  is  called  triangular,  quadrangular,  pen- 
tagonal, etc.,  according  as  its  base  is  a  triangle,  a 
quadrilateral,  a  pentagon,  etc. 

2.  A  right  prism  is  one  whose  principal  edges  are 
perpendicular  to  the  bases.  Any  other  prism  is  called 
oblique. 


84 


GEOMETRY. 


3.  A  cylinder  is   a   solid  described  by  the  revolu- 
tion of   a  rectangle  about  one  side,  which 
remains  fixed. 

The  fixed  side  is  called  the  axis  of  the 
cylinder.  The  opposite  side  describes  the 
convex  surface.  The  circles  described  by 
two  other  sides  are  called  the  bases. 

4.  A  prism  is  said  to  be  inscribed  in 
a    cylinder   when    its    bases    are    inscribed 

in   the  bases    of   the   cylinder,  and  its  principal  edges 
lie  in  the  convex  surface  of  the  cylinder. 

5.  The  HiGHT,  or  altitude,  of  a  prism  or  a  cylin- 
der, is  the  perpendicular  distance  between  the  planes 
of  its  bases. 


THEOREM    VI. 

The  convex  surface  of  a  right  prism  is  equal  to  the 
perimeter  of  its  base  multiplied  by  its  hight. 

Let  ABE  be  a  right  prism.  Since 
its  principal  edges  are  by  defini- 
tion perpendicular  to  the  bases,  any 
one  of  them  may  be  taken  as  the 
hight  of  the  prism.  Now,  AD  and 
BC,  being  both  perpendicular  to  the 
lower  base,  are  also  perpendicular  to 
AB,  which  they  meet  in  the  plane  of 
that  base  (Def.  1,  Sec.  XIII);  hence,  the  parallelo- 
gram ABCD  is  a  rectangle,  and  its  area  is  equal  to 
its  base  AB  multiplied  by  its  altitude  BC  (Cor.  1, 
Theo.  XVII,  B.  I).  In  the  same  manner  it  may  be 
shown  that  the  area  of  each  of  the  other  faces  com- 
posing the  convex  surface  is  equal  to  its  base  multi- 


BOOK    III 


85 


plied  by  the  altitude  of  the  prism.  Therefore,  the 
sum  of  their  areas  is  equal  to  the  sum  of  their  bases 
multiplied  by  the  common  altitude.  But  the  sum  of 
their  bases  constitutes  the  perimeter  of  the  base  of 
the  prism. 

Hence,  the  convex  surface,  etc. 


THEOEEM    VII. 

The  solidity  of  any  prism  is  equal  to  the  area  of  its 
base  multiplied  by  its  altitude. 

Let  ABCD  be  any  prism.  Now,  A| 
whatever  may  be  the  form  of  its 
base,  AB,  it  is  evident  that  it  may 
be  divided  into  an  indefinitely  large 
number  of  small  rectangles,  so  that 
the  remainder,  if  there  be  any,  shall 
be  less  than  any  assignable  quantity ;  and  the  base 
DC  may  be  divided  into  the  same  number  of  equal 
rectangles  having  their  sides  respectively  parallel  to 
the  sides  of  the  others.  Now,  each  rectangle  in  the 
lower  base,  and  its  corresponding  rectangle  in  the 
upper  base,  may  be  considered  as  the  opposite  bases 
of  a  small  parallelopiped  ;  and  the  prism  will  be  made 
up  of  such  parallelopipeds.  But  the  solidity  of  each 
of  these  will  be  equal  to  the  area  of  its  base  multi- 
plied by  its  altitude  (Cor.,  Theo.  Y),  which  is  the  same 
as  the  altitude  of  the  prism  :  hence,  the  sum  of  their 
solidities  will  be  equal  to  the  sum  of  their  bases  mul- 
tiplied by  their  common  altitude. 

That  is,  the  solidity  of  any  prism  is  equal,  etc. 


86 


GEOMETRY. 


THEOREM    Till. 

The  convex  surface  of  a  cylinder  is  equal  to  the  cir- 
cumference of  its  base  multiplied  by  its  altitude;  and 
its  solidity  is  equal  to  the  area  of  its  base  multiplied  by 
its  altitude. 

Let  iVBCD  be  a  cylinder,  having  A^ 
a  prism  inscribed  in  it  whose  base 
is  a  regular  polygon.  Now,  if  the 
number  of  sides  of  this  polygon  be 
indefinitely  increased,  its  perimeter 
will  ultimately  coincide  with  the  iC 
circumference  of  the  base  of  the 
cylinder.  Then,  also,  the  convex  surface  of  the  prism 
will  coincide  with  the  convex  surface  of  the  cylinder, 
and  the  solidity  of  the  prism  with  the  solidity  of  the 
cylinder.  But  the  convex  surface  of  the  prism  is 
equal  to  the  perimeter  of  the  base  multiplied  by  the 
altitude  (Theo.  VI),  and  its  solidity  is  equal  to  the 
area  of  the  base  multiplied  by  the  altitude  (Theo. 
VII). 

Therefore,  the  convex  surface  of  a  cylinder,  etc. 


SEC.  XVI.— PYRAMIDS    AND   CONES. 
DEFINITIONS. 

1.  A  pyramid  is  a  solid  bounded 
by  plane  faces,  of  which  one  is  any 
polygon,  and  the  others  triangles 
having  a  common  vertex.  The  poly- 
gon is  called  the  base.  The  triangles 
together  form  the  convex  or  lateral 
surface. 

Pyramids  are    called  triangular,  quadrangular,  pen- 


BOOK    III. 


87 


tagonal,  etc.,  according   as    their  bases   are  triangles, 
quadrilaterals,  pentagons,  etc. 

2.  A  regular  pyramid  is  one  whose  base  is  a  reg- 
ular polygon,  and  the  triangular  faces  are  equal  and 
isosceles. 

3.  A  cone  is  a  solid  described  by  the  revolution  of 
a  right-angled  triangle  about  one  of 
the  sides  containing  the  right  angle, 
which  side  remains  fixed.  The  fixed 
side  is  called  the  axis  of  the  cone. 
The  hypotenuse  describes  the  convex 
surface.  The  circle  described  by  the 
other  revolving  side  is  called  the  base. 

4.  The  altitude  of  a  pyramid  or  cone  is  the  per- 
pendicular distance  from  the  vertex  to  the  plane  of 
the  base. 

5.  The  slant  liiglit  of  a  regular  pyramid  is  the  per- 
pendicular let  fall  from  the  vertex  upon  the  base  of 
any  one  of  its  triangular  faces.  The  side  or  slant 
hight  of  a  cone  is  the  straight  line  drawn  from  the 
vertex  to  any  point  in  the  circumference  of  the  base. 

6.  A  FRUSTUM  of  a  pyramid  or  a  cone,  is  the  por- 
tion next  the  base  cut  off  by  a  plane  parallel  to  the 
base.  The  slant  hight  of  a  frustum  is  that  part  of 
the  slant  hight  of  the  whole  solid  which  lies  on  the 
frustum. 

7.  A  section  of  any  solid  is  the  surface  in  which 
it  is  divided  by  a  plane  which  passes  through  it. 


88 


GEOMETKT. 


THEOREM   IX. 

The  convex  surface  of  a  regular  pyramid  is  equal  to 
half  the  product  of  the  perimeter  of  the  base  by  the 
slant  hight. 

Let    ABODE    be    a    regular 

pyramid,*  of  Avhich  AF  is  the 
slant  hight.  The  area  of  the 
triangle  ACD  is  equal  to  half 
the  product  of  its  base  CD  into 
its  altitude,  which  is  the  slant 
hight.  Consequently  the  areas 
of  all  the  equal  triangles  com- 
posing the    convex    surface   are 

together  equal  to  half  the  product  of  the  sum  of  their 
bases  by  the  slant  hight.  But  the  sum  of  their  bases 
constitutes  the  perimeter  of  the  base  of  the  pyramid. 

Therefore,  the   convex    surface    of   a   regular   pyra- 
mid, etc. 

Schol.  1.  The  trapezoids  BF, 
CG,  etc.,  composing  the  convex 
surface  of  a  frustum  of  a  regular 
pyramid,  are  equal  to  each  other; 
for  they  are  the  differences  be- 
tween the  equal  triangles  AEF, 
AFGr,  etc.,  and  the  equal  tri- 
angles ABC,  ACD,  etc. 

Schol.  2.  Since  the  area,  of  the  trapezoid  BF  is 
equal  (Cor.  2,  Theo.  IX,  B.  II)  to  its  mean  breadth 
LM  multiplied  by  RS,  which  is  the  slant  hight  of  the 
frustum,  and  since  the  same  is  true  of  each  of  the 
other  trapezoids,  it  follows  that  the  convex  surface  of 
<i  frustum  of  a  regular  pyramid    is   equal   to  its  slant 


BOOK    III, 


89 


liiglit  multiplied  into  the  perimeter  of  a  middle  section 
between  its  two  bases. 


THEOKEM    X. 

Two   triangular  pyramids  of  equal  bases   and  alti- 
tudes arc  equivalent. 

Let  the  two  pyramids 
have  their  bases  in  the 
same  plane,  and  DC  equal 
to  BC.  Conceive  a  plane 
to  cut  the  two  solids 
parallel  to  the  plane  of 
their  bases,  making  the 
triangular    sections    FGH 

and  LMN.  Now,  since  the  vertices  R  and  S  are  by 
hypothesis  equidistant  above  the  plane  of  the  bases,  a 
third  plane  may  pass  through  these  two  points  par- 
allel with  the  other  two  planes  (Cor.  2,  Theo.  II). 
Then  the  lines  HL  and  RS,  being  the  intersections 
of  two  parallel  planes  by  a  third  plane  CRS,  will 
be  parallel  (Def.  4,  Sec.  XIII)  ;  hence,  the  triangles 
CRS,  CHL,  are  similar  (Cor.,  Theo.  V,  B.  II).  In 
the  same  manner  it  may  be  shown  that  the  triangles 
CRB,  HRG,  are  similar ;   also,  CSD  and  LSM. 

Now,  CR  :  HR  :  :  CB  :  HG. 

Also,  CS  :  LS  :  :  CD  :  LM. 

But  by  similarity  of  CRS  and  CHL,  we  have 

CR  :  HR  :  :  CS  :  LS. 
Hence,  by  equality  of  ratios  (Def.  0,  Sec.  X), 

CB  :  HG  :  :  CD  :  LM. 
But  CB  is  by  hypothesis  equal  to  CD ;  therefore,  HG 
is  equal  to  LM.     In  the  same  way  it  may  be  shown 
Evans    Geomelrrj. — 8 


90  GEOMETEY. 

that  the  other  sides  of  the  triangle  FHG  are  respect- 
ively equal  to  the  other  sides  of  the  triangle  LMN  ; 
and  these  triangles  are  consequently  equal  throughout. 
Hence,  at  equal  hights,  the  sections  parallel  to  the 
bases  are  equal ;  and  the  two  pyramids  may  bo  con- 
ceived to  be  applied  to  one  another  so  as  to  coincide 
at  all  equal  hights  successivel}r,  from  their  bases  to 
their  vertices.  They  are,  therefore,  equivalent. 
That  is,  two  triangular  pyramids,  etc. 

THEOKEM    XI. 

A  triangular  pyramid   is  one-third  of  a  triangular 
prism  of  the  same  base  and  altitude. 

Let   ABCDEF   be   a  triangular  a C 

prism.     Join    AF,    BF,    and   BD.  j         ,    1 

Now,  the    pyramid  BDEF,  cut  off  Umf 

by  the  plane  of  the  triangle  BDF,  h    /i,l 

is    equivalent    (Theo.     X)    to    the         /  ^p'j   \\ 
pyramid    FABC,    cut    off    by    the        |||f  '  '  W 
plane  of   the    triangle    ABF ;    for        ^Hh^ 
they  have   equal    bases,  DEF  and  jjj 

ABC  (Def.  1,  Sec.  XV),  and  the 
same  altitude,  namely  the  altitude  of  the  prism.  But 
the  pyramid  FABC  is  equivalent  to  the  third  pyramid 
BADF;  for  they  have  equal  bases  ADF  and  FCA 
(Cor.  1,  Theo.  XIII,  B.  I),  and  the  same  altitude, 
namely  the  perpendicular  distance  of  their  common 
vertex  B  above  the  plane  of  their  bases  ADFC. 
Hence,  the  pyramid  BDEF  is  one-third  of  the  prism 
ABCDEF. 

That  is,  a  triangular  pyramid  is  one-third,  etc. 

Cor.  1.     Hence,  the  solidity  of  a  triangular  pyramid 


BOOK    III. 


91 


is   equal  to   one-third  the  product  of  its  base  by  its 
altitude  (Thco.  VII). 

Cor.  2.  The  solidity  of  any  pyramid  ivhatever,  is 
equal  to  one-third  the  product  of  its  base  by  its  alti- 
tude; for,  by  dividing  its  base  into  triangles,  and 
passing  planes  through  the  lines  of  division  and  the 
vertex,  the  pyramid  can  be  divided  into  a  number  of 
triangular  pyramids  ;  and  the  sum  of  their  solidities 
will  be  equal  to  one-third  the  product  of  the  sum  of 
their  bases  by  their  common  altitude. 


THEOREM    XII. 

Tlie  convex  surface  of  a  cone  is  equal  to  half  the 
product  of  the  circumference  of  the  bast  by  the  sla?it 
hight;  and  its  solidity  is  equal  to  one-third  the  product 
of  the  area  of  the  base  by  the  altitude. 

Let  ABC  be  a  cone,  having  a 
regular  pyramid  inscribed  in  it. 
If  the  number  of  sides  of  the  poly- 
gon constituting  the  base  of  the 
pyramid  be  indefinitely  increased, 
its  perimeter  will  ultimately  coin- 
cide with  the  circumference  of  the 
base  of  the  cone.  Then,  the  slant 
hight  of  the  pyramid  will  be  equal 
to  the  slant  hight  of  the  cone,  and  the  convex  surface 
and  solidity  of  the  pyramid  to  the  convex  surface  and 
solidity  of  the  cone.  But  the  convex  surface  of  the 
pyramid  will  be  equal  to  half  the  product  of  the  peri- 
meter of  the  base  by  the  slant  hight  (Theo.  IX)  ;  and 
the  solidity  of  the  pyramid  will  be  equal  to  one-third 


92 


GEOMETRY. 


the   product  of  the   area  of  the  base  by  the  altitude 
(Cor.  2,  Tiieo.  XI). 

Therefore,  the  convex  surfaces  of  a  cone,  etc. 

Schol.  In  the  same  manner,  it  may  be  shown  that 
the  convex  surface  of  a  frustum  of  a  cone  is  equal  to 
the  product  of  the  slant  hight  into  the  circumference 
of  a  middle  section  between  the  two  bases  (Schol.  2, 
Theo.  IX).  And  as  this  will  hold  true  however  small 
the  upper  base  may  be,  it  will  hold  true  of  the  cone 
itself,  which  may  be  treated  as  a  frustum  whose  upper 
base  is  nothing. 


SEC.  XVII.— THE   SPHERE. 

* 

DEFINITIONS. 

1.  A  sphere  is  a  solid  which  may  be  described  by 
the  revolution  of  a  semicircle  around  its  diameter  as 
a  fixed  axis. 

The  semi-circumference  de- 
scribes the  convex  surface.  The 
center  is  the  middle  point  of 
the  axis.  The  radius  is  a 
straight  line,  from  the  center 
to  any  point  of  the  surface ; 
and  it  is  equal  to  the  radius 
of  the  semicircle.     A  diameter  is  a  double  radius. 

Schol.  As  the  semicircle  ADB  revolves  about  the 
axis  AB,  a  perpendicular,  as  DC,  let  fall  on  the  axis 
from  any  point  in  the  circumference,  will  describe  a 
circle. 

2.  A  great  circle    on   the    sphere    is    one   whose 


BOOK    III. 


93 


plane  passes  through  the  center.  Its  radius  is  the 
same  as  the  radius  of  the  sphere.  Its  circumference 
is  also  called  the  circumference  of  the  sphere.  Any 
other  circle  on  the  sphere  is  called  a  small  circle. 

3.  A  segment  of  a  sphere  is  any  portion,  as  ADEF, 
cut  off  from  the  solid  by  a  plane  passing  through  it. 

4.  A  frustum  of  a  cone  is  said  to  be  inscribed  in  a 
sphere,  when  the  circumferences  of  its  bases  lie  in  the 
surface  of  the  sphere. 


THEOREM   XIII. 

If  a  frustum  of  a  cone  be  inscribed  in  a  sphere,  its 
convex  surface  will  be  equal  to  the  altitude  of  the  frus- 
tum multiplied  by  the  circumference  of  a  circle,  whose 
radius  is  a  pcrpeyidicular  from  the  center  of  the  sphere 
to  the  slant  hight  of  the  frustum. 

Let  CD  and  GH  be  both  perpen- 
dicular to  the  axis  AB.  Then,  as 
the  semicircle  revolves  about  the 
axis,  describing  the  sphere,  the  trape- 
zoid DCGH  will  describe  a  frustum 
of  a  cone,  which  will  be  inscribed  in 
the  sphere.  From  the  center  K  let 
fall  the  perpendicular  KE,  on  the 
chord  CG ;  then  E  will  be  the  middle  point  of  CG 
(Theo.  XXV,  B.  I).  Draw  EF  perpendicular  to  AB, 
and  CI  perpendicular  to  GH. 

Now,  since  the  triangles  GCI,  KEF,  have  the  sides 
of  the  one  respectively  perpendicular  to  the  sides  of 
the  other,  they  are  similar  (Theo.  VI,  B.  II). 
Hence,  GC  :  CI  :  :  KE  :  EF. 


94  GEOMETKY. 

Multiplying  an  extreme  and  a  mean  equally  (Def.  5, 
Sec.  X), 

GC  :  CI  :  :  2  KEx  3.14159  :  2EFX3.14159. 
The  first  term  of  this  proportion  is  the  slant  hight  of 
the  frustum ;  the  second  is  its  altitude ;  the  third  is 
the  circumference  of  a  circle  whose  radius  is  KE  (Cor. 
1,  Theo.  XV,  B.  II) ;  the  fourth  is  the  circumference 
of  a  circle  'whose  radius  is  EF.  Therefore,  multi- 
plying extremes  and  means,  Ave  have  the  product  of 
the  slant  hight  into  the  circumference  of  a  middle 
section  of  the  frustum,  equal  to  the  product  of  its 
altitude  into  the  circumference  of  a  circle  whose 
radius  is  the  perpendicular  from  the  center  to  the 
slant  hight.  But  the  former  product  is  equal  to  the 
convex  surface  of  the  frustum  (Schol.,  Theo.  XII) ; 
consequently,  the  latter  product  is  equal  to  the  same. 
That  is,  if  a  frustum  of  a  cone  be  inscribed,  etc. 

Schol.  If  the  chord  CG  be  one  side  of  a  regular 
inscribed  polygon,  it  is  evident  that  KE  will  be  its 
apothegm  (Cor.  2,  Theo.  XXVIII,  B.  I). 

THEOKEM    XIV. 

TJie  surface  of  a  sphere  is  equal  to  the  product  of 
its  diameter  by  its  circumference. 

Let  ABCDEF  be  a  semicircle, 
having  the  half  of  a  regular  poly- 
gon inscribed  in  it.  As  the  semi- 
circle revolves  about  the  axis  AF, 
describing  the  sphere,  each  of  the 
trapezoids  GBCII,  IICDI,  etc.,  will 
describe  a  frustum  of  a  cone,  which 
will  be   inscribed  in  the  sphere.     The  convex  surface 


BOOK    III.  95 

of  each  of  these  frustums  will  be  equal  to  its  altitude 
multiplied  by  the  circumference  of  a  circle  whose 
radius  is  the  apothegm  of  the  polygon  (Schol.,  Theo. 
XIII).  Therefore,  the  sum  of  their  convex  surfaces 
will  be  equal  to  the  circumference  of  such  a  circle 
multiplied  by  the  sum  of  the  altitudes  Gil,  HI,  etc., 
that  is,  multiplied  by  AF,  the  diameter  of  the  sphere. 
Now  if  the  number  of  sides  of  the  semi-polygon  be 
indefinitely  increased, .  its  perimeter  will  ultimately 
coincide  with  the  semi-circumference,  and  its  apo- 
thegm with  the  radius  of  the  sphere.  Then,  the  sum 
of  the  convex  surfaces  of  the  frustums  will  be  equal 
to  the  surface  of  the  sphere,  and  the  circumference 
of  which  the  apothegm  is  radius  will  be  the  circum- 
ference of  the  sphere  (Def.  2). 

Therefore,  the  surface  of  a  sphere  is  equal,  etc. 

Cor.  1.  Since  the  circumference  is  equal  to  3.14159 
XD  (Cor.  1,  Theo.  XV,  B.  II),  it  follows  that  the  sur- 
face is  equal  to  3.141 59 XD2. 

Cor.  2.  The  convex  surface  of  any  segment  of  a 
sphere,  as  that  described  by  the  arc  ABC,  is  equal  to 
its  altitude  AH  multiplied  by  the  circumference  of  the 
sphere. 

THEOEEM   XV. 

Tlie  solidity  of  a  sphere  is  equal  to  its  surface  mul- 
tiplied by  one-sixth  of  its  diameter. 

The  sphere  may  be  considered  as  made  up  of  in- 
finitely small  pyramids,  whose  bases  together  form  the 
surface  of  the  sphere,  and  whose  common  vertex  is  at 
the  center.  Now,  the  solidity  of  each  of  these  pyra- 
mids will  be  equal  (Cor.  2,  Theo.  XI)  to  the  product 


96  GEOMETRY. 

of  its  base  by  one-third  of  its  altitude,  that  is,  by 
one  third  the  radius  of  the  sphere.  Hence,  the  sum 
of  their  solidities  will  be  equal  to  the  sum  of  their 
bases  multiplied  by  one-third  of  the  radius,  or  one- 
sixth  of  the  diameter. 

That  is,  the  solidity  of  a  sphere  is  equal,  etc. 

Cor.     Since    the    surface    of   a    sphere   is    equal   to 
8.141 59  XD2,  it   follows    that   its    solidity  is    equal    to 

D 

3.14159xD2X— ,  which,  by  reduction,  becomes  .5236X 
6 

D3.     That   is,   the  solidity  of  a  sphere  is  equal  to  the 

cube  of  its  diameter  multiplied  by  .5286. 


SUPPLEMENT. 


SEC.  XVIII.— MISCELLANEOUS    EXAMPLES. 

1.  What  is  the  side  of  a  square  inscribed  in  a  circle 
whose  diameter  is  5  feet? 

2.  Prove  that  the  side  of  a  square  is  to  its  diag- 
onal as  1  to  the  square  root  of  2. 

3.  Prove  that  the  area  of  a  square  described  about 
a  circle  is  double  the  area  of  a  square  inscribed  in 
the  same  circle. 

4.  "What  is  the  altitude  of  an  equilateral  triangle 
whose  side  is  12  feet  ? 

5.  If  the  altitude  of  a  cone  is  9  feet,  and  the  di- 
ameter of  its  base  5  feet,  4  inches,  what  is  its  convex 
surface  ?     Its  solidity  ? 

6.  What  is  the  surface  of  a  sphere  avIiosc  diameter 
is  7  feet?     Its  solidity? 

7.  If  the  altitude  of  a  prism  is  5  feet,  and  the  area 
of  its  base  18  square  feet,  what  is  its  solidity? 

8.  What  is  the  solidity  of  a  cube  whose  edge  is  G 
feet?     Its  surface? 

9.  What  is  the  lateral  surface  of  a  regular  pyramid 
whose  slant  hiiiht  is  15  feet,  and  the  base  80  feet 
square? 

Evans   Geometry. — 9  (27) 


98  GEOMETRY.  ■ 

10.  What  is  the  solidity  of  a  pyramid  whose  alti- 
tude is  72,  d  the  sides  of  whose  base  are  24,  24, 
and  40  ? 

11.  If  the  altitude  of  a  cylinder  is  7,  and  the 
radius  of  its  base  2,  what  is  its  whole  surface  ?  Its 
solidity  ? 

12.  Prove  that  a  circle  described  on  the  hypotenuse 
of  a  right-angled  triangle  as  diameter,  is  equivalent  to 
the  sum  of  the  circles  described  on  the  other  two 
sides. 

13.  Prove  that  if  a  perpendicular  be  let  fall  from 
the    right    angle    of    a    right-angled    triangle    on    the 

©  ©  ©  ©  © 

hypotenuse,  either  of  the  two  sides  containing  the 
right  angle  will  be  a  mean  proportional  between  the 
hypotenuse  and  the  adjacent  part  of  the  hypotenuse. 

14.  If  from  the  top  of  a  mountain  2.006  miles 
high  a  vessel  can  be  seen  in  the  horizon  126  miles 
off,  what  is  the  diameter  of  the  earth  ?  (Theo.  XIII, 
B.  II). 

15.  If  the  earth's  diameter  be  7912  miles,  what  is 
its  circumference  ?     Its  surface  ?     Its  solidity  ? 

16.  If  the  circumference  of  the  moon  be  6783  miles, 
what  is  its  diameter? 

17.  Prove  that  the  solidities,  S  and  s,  of  any  two 
spheres  are  to  each  other  as  the  cubes  of  their  diame- 
ters, D  and  d. 

18.  The  diameter  of  the  sun  is  112  times  that  of  the 
earth  ;  what  arc  the  relative  solidities  of  the  two  bodies  ? 


SUPPLEMENT.  99 


SEC.  XIX.— APPLICATIONS  OF  ALGEBRA. 

For  those  acquainted  with  the  elements  of  hoth  branches, 
a  few  examples  of  the  application  of  Algebra  to  Geometry 
are  subjoined.  The  method  is  specially  adapted  to  the 
solution  of  that  class  of  problems  where  certain  parts  of  a 
figure  are  aiven,  from  which  to  determine  others  ;  but  it 
may  also  be  used  for  abbreviating  long  demonstrations  of 
theorems. 

1.  If  a  perpendicular  be  let  fall  from  the  vertex  of 
a  triazigle  upon  its  base,  the  sum  of  the  parts  of  thu 
base  is  to  the  sum  of  the  other  two  sides  as  the  dif- 
ference of  the  latter  is  to  the  difference  of  the  former. 

Proof.  Let  BD  be  the  per- 
pendicular. Now,  BC2—  CD-  = 
BD2  (Cor.  2,  Theo.  XIX,  B.  I).  / 

Also,         AB2— AD2=BD2.  / 

Therefore,  A       d  ~~c 

BC2— CD2=AB2— AD2. 

Transposing,     BC2— AB2=CD2—  AD2. 

Factoring, 

(BC-fAB)  (BC— AB)  =  (CD+AD)  (CD— AD). 
Hence,  resolving  into  a  proportion,  Ave  have 

CD-fAD  :  CB+AB  :  :  CB— AB  :  CD— AD. 

2.  The  square  of  the  side  opposite  to  any  acute 
angle  in  a  triangle  is  equal  to  the  sum  of  the  squares 
of  the  other  sides,  minus  twice  the  product  of  the 
base  by  the  distance  from  the  acute  angle  to  the  foot 
of  the  perpendicular  let  fall  from  the  vertex  on  the 
base,  or  the  base  produced. 


100 


GEOMETRY. 


Proof.     For   the    case   where 

the    perpendicular   falls    on    the 

base  produced,  take  the  annexed 

figure ;    for  the  other  case,  take 

the  preceding  one. 

y  y        a 

Now,    AB2=BD4-AD2  (Theo.  XIX,  B.  I). 

But,  AD=CD— AC  (or,  AC— CD). 

Hence,      AD-  =  CD--{-AC2— 2  CDx  AC. 

Substituting  this  in  the  first  equation,  we  have 

AB2 =BD'-+  CD-+ AC2— 2  CD  X  AC. 

But  BD2+CD2=BC2;  whence, 

AB-=BC-+AC2— 2  CDX  AC. 

3.  The  square  of  the  side  opposite  an  obtuse  angle 
in  a  triangle  is  equal  to  the  sum  of  the  squares  of 
the  other  sides,  plus  twice  the  product  of  the  base  by 
the  distance  from  the  obtuse  angle  to  the  foot  of  the 
perpendicular,  let  fall  from  the  vertex  on  the  base 
produced. 

For  proof  use  the  last  figure. 

4.  Given  the  base  and  the  sum  of  the  two  other 
sides  of  a  right-angled  triangle,  to  find  the  hypote- 
nuse and  perpendicular. 

Solution.  Represent  the  base  by  b  and  the  sum 
of  the  other  sides  by  s.  Also,  represent  the  perpen- 
dicular by  x,  then  the  hypotenuse  Avill  be  s — x. 

Now,  V+x2={8—xf  (Theo.  XIX,  B.  I). 

Hence,  by  reduction,  we  have, 

O  TO 

8~ — 0' 


X  =  - 


2s 


SUPPLEMENT. 


101 


5.  Given  the  base  and  altitude  of  a  triangle,  to  find 
the  side  of  the  inscribed  square. 

Solution.  Represent  the 
base  AC  by  b,  the  altitude 
BH  by  h,  and  the  side  of  the 
inscribed  square  by  x.  Then 
BI  will  be  represented  by 
h — x. 

Now,  by  similarity  of  triangles, 

h  :  b  :  :  Ji — x  :  x. 
Hence,  hx=bh — bx. 

bh 


And, 


x  = 


b+h 


6.  If  two  parallel  chords  in  a  circle  be  96  and 
(30,  and  the  distance  between  them  26,  what  is  the 
diameter  ? 

Solution.  Draw  the  di- 
ameter AF  perpendicular  to 
the  chords,  and  it  will  bisect  j/ 
them  (Theo.  XXV,  B.  I). 
Represent  AH  by  x  and  GF 
by  y.  Here,  since  we  have 
introduced  two  unknown  quan- 
tities, we  must  have  two  independent  equations. 

Now,  x(26+y)=(SQ)2  (Theo.  XII,  B.  II). 
And  04-26)7/=(48)2. 

Finding  x  and  y  from  these  equations,  and  adding 
their  sum  to  26,  we  shall  have  the  diameter =100. 

7.  Given  the  diagonal  and  perimeter  of  a  rectangle, 
to  find  the  sides. 


102  GEOMETRY. 

8.  In  a  triangle,  given  the  base  b,  the  altitude  h, 
and  the  ratio  of  the  other  two  sides,  as  m  to  n,  to  find 
the  sides. 

Represent  the  sides  by  mx  and  nx. 

9.  Given  the  base  and  altitude  of  any  triangle 
and  the  difference  of  the  other  two  sides,  to  find 
the  sides. 

Let  x  stand  for  the  half  sum  of  the  two  sides, 
and  d  for  their  half  difference,  then  the  sides  will  be 
represented  by  x-\-d  and  x — d. 

10.  Given  the  base  and  the  other  two  sides  of  a 
triangle,  to  find  the  altitude. 

11.  Given  the  area  of  a  rectangle  inscribed  in  a 
given  triangle,  to  find  the  sides  of  the  rectangle. 

12.  Given  the  radius  of  a  circle,  to  find  the  side 
of  the  inscribed  equilateral  triangle. 


0^  THE 


THE  MOST  POPULAR 


SCHOOL-BOOKS  IN  THE  WEST. 


INDORSED   BY 


THE  BEST  MEN  IN  THE  COUNTRY. 


The  attention  of  Parents,  Teachers,  and  School-Officers 
is  respectfully  invited  to  the  following  recommendations 
of  the  Eclectic  Educational  Series,  from  a  large 
number  of  the  leading  educators  of  the  day,  including 
Superintendents  of  Public  Instruction  of  nearly  all  the 
Western  States. 

These  testimonials,  together  with  others  from  gentlemen 
of  the  highest  educational  positions,  and  of  large  practical 
experience  in  the  school-room,  show  conclusively  that  the 
great  and  abiding  popularity  of  the  Eclectic  Series  is 
founded  upon  their  rare  intrinsic  merit,  and  their  admir- 
able adaptation  to  the  wants  of  schools. 


OHIO. 

From  Hon.  Anson"  Smtth,  former  Slate  Supt.  of  Public  Inst,  Ohio. 

McGuffey's  Old  Eclectic  Readers  I  esteemed  as  among 
the  very  best  works  of  the  kind ;  but  the  New  are  certainly  a 
decided  improvement  upon  the  Old.  I  know  of  no  others 
which  I  could  more  earnestly  and  honestly  indorse. 

I  have  examined  with  care  the  new  editions  of  Ray's 
Arithmetics,  and  am  greatly  pleased  with  them.  The  Pri- 
mary is  certainly  a  very  decided  improvement.  I  have  seen 
none  so  well  adapted  for  a  text-book  in  the  elements  of  Men-* 
tal  Arithmetic.     The  Intellectual  is  an  admirable  work. 


2  POPULAR  SCHOOL-BOOKS. 

I  know  of  no  other  that  embodies  so  systematic,  complete, 
and  thorough  a  course  of  discipline  in  this  useful  branch  of 
study.  Ray's  Practical  Arithmetic  needs  no  praise.  It 
is  its  own  commendation. 

Pinneo's  Series  of  Grammars  I  esteem  as  amonjr  the  best 
text-books  extant,  for  guiding  the  learner  to  a  knowledge  of 
the  correct  use  of  our  language.  The  definitions  are  clear 
and  exact ;  the  rules,  simple  and  comprehensive,  and  the 
whole  plan  and  arrangement,  well  adapted  to  achieve  their 
purpose.  Anson  Smyth. 

INDIANA. 
From  Hox.  Miles  J.  Fletcher,  former  Stale  Supt.  of  Public  Inst,  Ind. 

The  public  sentiment,  as  expressed  in  Indiana  by  the  al- 
most universal  use  of  the  Eclectic  Educational  Series 
of  School  Books,  embracing  McGuffey's  New  Readers  and 
Speller,  Ray's  Series  of  Arithmetics  and  Algebras,  and  Pin- 
neo's Grammars,  was  sufficient  in  itself  to  induce  the  State 
Board  of  Education  to  adopt  them. 

In  addition  to  this,  by  careful  examination,  I  am  well  sat- 
isfied that  their  true  intrinsic  and  relative  merit  entitles  them 
to  such  recommendation.  They  are  printed  on  good,  firm, 
substantial  white  paper,  are  durably  bound,  and  of  unrivaled 
cheapness.  Miles  J.  Fletcher. 

ILLINOIS. 
From  Hox.  N.  Batemax,  State  Supt.  of  Public  Inst.,  Illinois. 

No  series  of  books  has  ever  obtained  so  many  voices  of 
approval  from  teachers  as  McGuffey's  Eclectic  Readers. 
Certainly  no  other  series  has  been  so  popular  throughout  the 
West.  We  unhesitatingly  say  that  we  know  of  no  letter  books, 
and  should  not  take  the  trouble  to  look  for  any.  The  print- 
ing is  beautiful,  the  paper  very  fine,  and  the  binding  good, 
and  McGuffey's  Readers  are  proverbially  cheap. 

Ray's  Arithmetics  have  deservedly  shared  in  the  popu- 
larity of  the  Eclectic  Series.  The  Higher  Arithmetic  is  better 
than  any  other  that  wc  know  to  be  used  in  this  country. 
Ray's  Algebras  are  clear,  full,  and  comprehensive. 


POPULAR  SCHOOL-BOOKS.  8 

If  I  were  to  select  from  the  different  Series  of  School- 
Books  now  used  in  the  West,  those  which,  upon  the  whole,  I 
deemed  best  adapted  to  the  Public  District  Schools  of  Illi- 
nois, I  should  choose  McGuffey's  New  Eclectic  Series  of 
Readers,  and  Kay's  Series  of  Arithmetics  and  Algebras,  as 
combining  more  excellences  and  fewer  defects  than  any  other 
series  with  which  I  am  acquainted.  The  preference  herein 
expressed  is  the  result  of  actual  use,  in  my  own  schools,  of 
some  live  or  six  different  Series  of  Readers,  Arithmetics,  and 
Algebras,  as  well  as  a  diligent  comparison  of  all  the  leading 
series  which  have  been  published  in  the  last  twenty  years. 

Newton  Bateman. 

We  heartily  concur  in  the  foregoing  recommendation. 
[Signed.]  C.  Goudy, 

Member  of  the  Illinois  State  Board  of  Education,  ami 
Chairman  of  Committee  on  Text-Books  iu  the  Normal  University. 

S.  W.  MOULTON, 

President  of  the  Illinois  State  Board  of  Education. 

Henry  Wing, 

Member  of  the  Illinois  State  Board  of  Education, 

Richard  Edwards, 

President  of  the  Illinois  State  Normal  University. 


IOWA. 

Fro)n  Hon.  Oran  Faville,  former  State  Supt.  of  Public  Inst.,  Iowa. 

Having  recently  re-examined  the  Eclectic  Educational 
Series  of  Common  School  Books,  I  am  fully  confirmed  in 
the  opinion  that  they  are  the  best  Series,  on  the  whole,  now 
in  use  in  the  West.  Their  remarkable  popularity,  and  the 
continued  attachment  manifested  for  them  by  practical  Edu- 
cators, give  evidence  both  of  their  intrinsic  worth,  and  of 
their  adaptation  to  the  place  designed  for  them. 

Without  specifying  further,  I  will  say  that  McGuffey's 
New  Eclectic  Series  of  Readers,  Speller,  and  Primary 
School  Charts,  Pinneo's  Series  of  Grammars,  and  Ray's 
Series  of  Arithmetics  and  Algebras,  are  tin  surpassed  by 
any  similar  Series  with  which  I  am  acquainted.  I  therefore 
recommend  their  continued  use  in  our  State. 

Oran  Faville. 


4  POPULAR  SCHOOL-BOOKS. 

From  Hon.  H.  A.  Wiltse,  former  State  Supt.  of  Public  Inst.,  Iowa. 

I  have  long  been  acquainted  with  the  Eclectic  Educa- 
tional Series,  embracing  McGuffey's  New  Readers, 
Speakers,  and  Speller,  Ray's  Arithmetics  and  Alge- 
bras, and  Pinneo's  Grammars.  I  have  seen  this  Series 
tested  by  years  of  practice,  have  had  opportunity,  extending 
through  years,  of  comparing  its  results,  and  know  that  it  is 
the  best  Series  published.  The  Schools  that  have  used  McGuf- 
fey's Speller,  Readers,  and  Speakers,  are,  under  equal  instruc- 
tion, in  advance  of  Schools  where  these  books  have  not  been 
used.  H.  A.  Wiltse. 


WISCONSIN. 

From  Hon.  J.  L.  Pickard,  former  State  Supt.  of  Public  Inst.,   Wis.; 
at  present,  Supt.  of  Chicago  Public  Schools. 

The  books  I  have  recommended  below,  [McGuffey's  New 
Readers,  Ray's  Arithmetics,  Pinneo's  Grammar,  and  White's 
Class-Book  of  Geography,]  are  such  as  commend  themselves 
to  my  judgment.  I  would  advise  their  adoption  in  all  schools 
where  no  uniformity  at  present  exists.  J.  L.  Pickard. 


MICHIGAN. 

From   Edward    Olney,   Professor    of   Mathematics  in    the   State 
University  of  Michigan. 

University  of  Michigan,  July  2d,  1866. 
I  am  using  Ray's  Algebra  (Higher)  because  I  know  of  no 
better.  I  have  a  tolerable  familiarity  with  the  various  Al- 
gebras published  in  our  country,  and  use  Ray's  because  of 
its  dear,  condensed  style,  copious  and  diversified  examples, 
and  the  judicious  selection  of  topics  to  be  presented.  In 
short',  lor  a  text-book  in  our  Colleges  and  Higher  Seminaries, 
I  think  it  not  only  has  no  superior  but  no  equal  published  in 
our  country ;  and  hence  I  use  it  till  I  find  a  better. 

Edward  Olney. 


POPULAR  SCHOOL-BOOKS.  5 

MINNESOTA. 

From  Hon.  B.  F.  Crary, /ormer  State  Supt.  of 'Public  Inst.,  Minn. 

I  have  examined  McGuffey's  New  Eclectic  Readers, 
and  have  no  hesitation  in  saying  that  they  are  superior  to  any 
similar  text-books  that  have  come  under  my  observation. 
The  standard  of  morals  and  taste  in  the  Readers  is  very 
high,  and  in  their  low  price  and  beautiful  printing  and  bind- 
ing, they  distance  all  competition.  B.  F.  Crary. 


MIS30UEL 

From  Hon.  James  H.  Robinson,  late  State  Supt.  of  Public  Inst.,  Mo. 
After  careful  examination  of  their  intrinsic  and  compara- 
tive merits,   I   cordially  recommend   for  use  in  the    Public 
Schools  of  Missouri,  the  following  Text-Books : 


McGuffey's  Primary  School  Charts, 
McGuffey's  New  Eclectic  Speller, 
McGuffey's  New  Eclectic  Readers, 
Kidd's  Elocution  and  Vocal  Culture, 


Ray's  Series  of  Arithmetics, 
Ray's  Series  of  Algeukas, 
Ray's  Plane  and  Solid  Geometry, 
Pinneo's  Series  of  Grammars. 


No  other  similar  books  have  been  so  largely  used,  and  are 
so  generally  popular  as  Ray's  Mathematics  and  McGuffey's 
Readers,  and  I  believe  them  to  be  better  adapted  to  the 
wants  of  our  Schools  than  any  other  works  of  the  kind  pub- 
lished. James  IT.  Robinson. 


KANSAS. 

From  Hon.  Wm.  R.  Griffith,  former  State  Supt.  of  Public  Inst.,  Kan. 

I  recommend  McGuffey's  New  Eclectic  Series  of 
Readers,  Speakers,  and  Speller,  and  Ray's  Series  of 
Arithmetics  and  Algebras  to  the  favorable  consideration 
of  the  Teachers  of  our  Public  Schools.  These  works  possess 
real  merit,  and  I  trust  they  will  be  approved  by  the  citizens 
of  the  State  generally. 

I  have  also,  after  careful  examination,  concluded  to  rec- 
ommend Pinneo's  Series  of  Grammars.  I  have  endeav- 
ored to  examine  the  most  popular  works  on  the  subject  of 


C  POPULAR  SCHOOL-EOOKS. 

Grammar,  as  a  teacher  rather  than  as  a  critic,  and,  in  sd 
doing,  have  been  compelled  to  give  my  preference  to  Pinneo's. 
The  early  introduction  of  analysis,  and  the  abundant  black- 
board exercises  provided,  make  Pinneo's  Grammars  very  prac- 
tical works.  Wm.  R.  Griffith. 

PENNSYLVANIA. 

From  Prof.  J.   P.   \Vickeusiiam,  State  Supt  of  Public  Inst,  Pain.: 
late  Principal  of  the  State  Nonnal  School 

\Vc  use  Ray's  Algebras  in  this  institution,  and  consider 
them  the  best  text-books  on  the  subject  of  Algebra  published 
in  this  country.  J-  P-  Wickersham. 

From  Hon.  T.  II.  Burrowes,  former  State  Supt.  of  Public  Inst,  Penn. 
It  is  needless  to  call  the  attention  of  experienced  teachers 
to  Ray's  books— Arithmetics,  Algebras,  and  Geometry.  They 
have  stood  the  test  of  long  trial,  and  are  annually  increas- 
ing in  their  circulation.  Ray's  Geometry  is  one  of  the  best 
Elementary  Treatises  on  this  subject,  characterized  by  logical 
accuracy  of  its  definitions,  simplicity  of  its  methods,  and  the 
rational  and  consecutive  arrangement  of  the  subject  matter 
of  which  it  treats.  T.  H.  Burrowes. 

WEST  VIRGINIA. 
From  Hon.  Wm.  R.  White,  State  Supt.  of  Public  Inst.,  West  Va. 

In  compliance  with  the  Second  Section  of  the  School  Law, 
the  following  books  are  prescribed  to  be  used  in  the  Free 
Schools  throughout  West  Virginia:  McGuffey's  New  Ec- 
lectic Series  of  Readers,  Charts,  and  Speller  ;  Kill's 
Elocution,  Ray's  Series  of  Arithmetics  and  Algebras, 
Evans'  School  Geometry,  and  Pinneo's  Grammars  and 
Guide  to  Composition. 

Dr.  Pinneo's  books  are  eminently  adapted  to  aid  the  teacher 
as  well  as  the  learner.  Those  who  use  them,  following  the 
Hints  to  Instructors  and  Suggestions,  will  find  Grammar  robbed 
of  many  of  the  difficulties  usually  attending  the  study  of  this 
science.  \Wm.  R.  White- 

UNIVERSITY  J 

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